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All questions of Periodic Table for Chemistry Exam

Which of the following bond is strongest
  • a)
    C – C
  • b)
    C – H
  • c)
    C – O
  • d)
    None
Correct answer is option 'B'. Can you explain this answer?

Vikram Kapoor answered
Because of small differences in electronegativities, the C−H bond is generally regarded as being non-polar. The real reason here is atomic size. The hydrogen atom is much smaller than the carbon atom. Smaller bonds lead to higher bond energy; therefore C−H bond has higher bond enthalpy than the C−C bond.
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First ionization energy is the lowest with:
  • a)
    Lead
  • b)
    Carbon
  • c)
    Silicon
  • d)
    Tin
Correct answer is option 'D'. Can you explain this answer?

Asf Institute answered
Ionization energy decreases down the group. As charge density is minimum for tin so first ionization energy is lowest for tin.

If the Aufbau principle had not been followed, Ca (Z = 20) would have been placed in the:
a) s-block
b) p-block
c) d-block
d) f-block
The correct answer is option 'C'. Can you explain this answer?

Aditya Deshmukh answered
Electronic configuration of Ca is 1s2 2s2 2p6 3s2 3p6 4s2. 
Filling of 3d orbital starts after the filling of 4s orbitals. If the aufbau principle had not been followed, the filling of 3d orbital would have been prior to filling of 4s orbital and the electronic configuration of Ca would have been   1s2 2s2 2p6 3s2 3p6 3d2 and it belongs to d block.

The third ionization energy is maximum for:
  • a)
    Boron 
  • b)
    Phosphorus
  • c)
    Aluminium
  • d)
    Nitrogen
Correct answer is option 'D'. Can you explain this answer?

Shivam Sharma answered
Explanation:

Ionization energy is the energy required to remove an electron from an atom or ion. The third ionization energy refers to the energy needed to remove the third electron.

Looking at the electronic configurations of the given elements:

1. Boron (B): 1s² 2s² 2p¹
2. Phosphorus (P): 1s² 2s² 2p⁶ 3s² 3p³
3. Aluminium (Al): 1s² 2s² 2p⁶ 3s² 3p¹
4. Nitrogen (N): 1s² 2s² 2p³

For Nitrogen, the electronic configuration of the outer shell is a half-filled configuration (2s² 2p³). This configuration is considered stable, as half-filled orbitals are more stable due to their symmetric arrangement and better shielding effect.

When we try to remove the third electron from nitrogen (third ionization), we are disturbing the stable half-filled configuration, which would require more energy. Thus, the third ionization energy is maximum for nitrogen among the given elements.

The correct order of ionic size of N3–, Na+, F–, Mg2+ and O2– is:
  • a)
    Mg2+ > Na+ > F– > O2– < N3–
  • b)
    N3– < F– > O2– > Na+ > Mg2+
  • c)
    Mg2+ < Na+ < F– < O2– < N3–
  • d)
    N3– > O2– > F– > Na+ < Mg2+
Correct answer is option 'C'. Can you explain this answer?

Siddharth Banerjee answered
If the ions derived from different atoms are isoelectronic species, then they all have same number of electrons in their electronic shells and will have got same electronic configuration but their nuclear charge will differ because of their difference in number of protons in the nucleus. With increase in number of protonsin the nucleus the electrons are more attracted towards nucleus thereby causing the decrease in ionic radius. On this principle our problem will be solved
The given ions are
  • 7N3-→no. of proton=7 and no of electron=10
  • 8O2-→no. of proton=8 and no of electron=10
  • 9F-→no. of proton=9 and no of electron=10
  • 11Na+→no. of proton=11 and no of electron=10
  • 12Mg2+→no. of proton=12 and no of electron=10
Hence the increasing order of ionic radius is
12Mg2+< 11Na+< 9F-< 8O2-< 7N3-
For isoelectronic species lower the nuclear charge higher the radius

Incorrect order of ionic size is:
  • a)
    La3+ > Gd3+ > Eu3+ > Lu3+
  • b)
    V2+ > V3+ > V4+ > V5+
  • c)
    Tl+ > In+ > Sn2+ > Sb3+
  • d)
    K+ > Sc3+ > V5+ > Mn7+
Correct answer is option 'A'. Can you explain this answer?

Pooja Choudhury answered
In a given lanthanide series , ionic radii decreases due to lanthanoid contraction . So size of Eu3+ should be larger than Gadolinium ion Gd3+
Hence option A is the answer.

Which of the following ions is most unlikely to exist?
  • a)
    Li–
  • b)
    Be–
  • c)
    B–
  • d)
    F–
Correct answer is option 'B'. Can you explain this answer?

Swara Reddy answered
Be- ions is most unlikely to exist as Be has higher tendency to loose electron and Be has full filled configuration.

The first ionization potential in electron volts of nitrogen and oxygen atoms are respectively given by:
a)14.6, 13.6
b)13.6, 14.6
c)13.6, 13.6
d)14.6, 14.6
The correct answer is option 'A'. Can you explain this answer?

Rashi Choudhury answered
Nitrogen has higher ionization energy than Oxygen because it has a stable half-filled electronic configuration as shown below:
7 N - (1s)2(2s)2(2px)1(2py)1(2pz) and
8 O - (1s)2(2s)2(2px)2(2py)1(2pz)1

In which of the following pair, both the species are isoelectronic but the first one is large in size than the second?
  • a)
    S2–, O2–
  • b)
    Cl–, S2–
  • c)
    F–, Na+
  • d)
    N3–, P3–
Correct answer is option 'C'. Can you explain this answer?

Bhavana Dasgupta answered
F- and Na+ both have 10 electrons .Atomic no of F is 9 and for Na it is 11.so for F-, 9 proton is pulling 10 electrons and for Na+,11 proton is pulling 10 electrons . So attractive force is more for Na+ compared to F-. So radius of F- is more than Na+.

The correct values of ionization enthalpies (in kJ mol–1) of Si, P, Cl and S respectively are:
  • a)
    786, 1012, 999, 1256
  • b)
    1012, 786, 999, 1256
  • c)
    786, 1012, 1256, 999
  • d)
    786, 999, 1012, 1256
Correct answer is option 'C'. Can you explain this answer?

Mihir Singh answered
-1) for the first 20 elements are:

1. Hydrogen - 1312
2. Helium - 2372
3. Lithium - 520
4. Beryllium - 899
5. Boron - 801
6. Carbon - 1086
7. Nitrogen - 1402
8. Oxygen - 1314
9. Fluorine - 1681
10. Neon - 2081
11. Sodium - 496
12. Magnesium - 738
13. Aluminum - 578
14. Silicon - 786
15. Phosphorus - 1012
16. Sulfur - 1000
17. Chlorine - 1251
18. Argon - 1521
19. Potassium - 419
20. Calcium - 590

What is the atomic number of the element with the maximum number of unpaired 4p electron:
  • a)
    33
  • b)
    26
  • c)
    23
  • d)
    15
Correct answer is option 'A'. Can you explain this answer?

Shivam Sharma answered
33, because p block has 6 elements per period.

Max unpaired is halfway, so 3rd.

3rd element of 4th period p-block is 33.

The correct order of increasing electron affinity of the following elements is:
  • a)
    O < S < F < Cl
  • b)
    O < S < Cl < F
  • c)
    S < O < F < Cl
  • d)
    S < O < Cl < F
Correct answer is option 'A'. Can you explain this answer?

Amar Chawla answered
Order of Increasing Electron Affinity:

Electron affinity refers to the tendency of an atom to attract an electron towards itself. The correct order of increasing electron affinity of the given elements is as follows:

a) O S F Cl

Explanation:

The correct order of increasing electron affinity can be explained as follows:

- Oxygen (O) has the least electron affinity as it has a small atomic size and a stable electronic configuration. It requires less energy to remove an electron from its outermost shell, making it less likely to attract an additional electron.
- Sulfur (S) has a slightly higher electron affinity than oxygen due to its larger atomic size and lower effective nuclear charge. This makes it easier for sulfur to attract an electron towards itself.
- Fluorine (F) has a higher electron affinity than sulfur due to its smaller atomic size and high effective nuclear charge. It has a strong attraction towards electrons and tends to form stable anions.
- Chlorine (Cl) has the highest electron affinity among the given elements due to its larger atomic size and lower effective nuclear charge. It has a stronger attraction towards electrons than fluorine and forms stable anions.

Therefore, the correct order of increasing electron affinity of the given elements is O < s="" />< f="" />< cl.="" />

How does the energy gap between successive energy levels in an atom vary from low to high n values:
  • a)
    All energy gaps are the same
  • b)
    The energy gap decreases as n increases
  • c)
    The energy gap increases as n increases
  • d)
    The energy gap changes unpredictably as n increases
Correct answer is option 'B'. Can you explain this answer?

Ishani Dasgupta answered
The further away an electron is from the nucleus, the less force it feels from the electron, so the less energy is needed to “pop it off” the atom. The value of the energy level is exactly this amount of energy  so the smaller it is, the smaller the difference with neighboring levels will be.

A, B and C are hydroxy-compounds of the elements X, Y and Z respectively. X, Y and Z are in the same period of the periodic table. A gives an aqueous solution of pH less than seven. B reacts with both strong acids alkalis. C gives an aqueous solution which is strongly alkaline.
Which of the following statements is/are true?
(I) The three elements are metals.
(II) The electronegativities decrease from X to Y to Z
(III) The atomic radius decreases in the order X, Y and Z
(IV) X, Y and Z could be phosphorus aluminium and sodium respectively.
  • a)
    I, II, III only correct
  • b)
    I, III only correct
  • c)
    II, IV only correct
  • d)
    II, III IV only correct 
Correct answer is 'C'. Can you explain this answer?

Pooja Choudhury answered
  • The electronegativities decrease from X to Y to Z.
  • X, Y and Z could be phosphorus aluminium and sodium respectively.

Which one of the following statements is incorrect:
  • a)
    Greater is the nuclear charge, greater is the electron gain enthalpy
  • b)
    Nitrogen has almost zero electron gain enthalpy
  • c)
    Electron gain enthalpy decreases from fluorine to iodine in the group
  • d)
    Chlorine has highest electron gain enthalpy
Correct answer is option 'C'. Can you explain this answer?

Shivam Sharma answered
The incorrect statement is:
Electron gain enthalpy decreases from fluorine to iodine in the group.
Explanation:
Electron gain enthalpy is the energy released when an electron is added to an isolated gaseous atom. In general, electron gain enthalpy becomes more negative (i.e., more energy is released) as we move across a period from left to right and up a group from bottom to top in the periodic table. This is due to the increasing effective nuclear charge and decreasing atomic size, which makes it easier for the atom to attract and hold an additional electron.
However, statement 3 is incorrect because electron gain enthalpy actually increases (becomes less negative or more positive) as we move down a group, such as from fluorine to iodine. This is because, as we move down a group, the atomic size increases and the effective nuclear charge experienced by the incoming electron decreases. This makes it less favorable for the atom to gain an electron, resulting in a less negative (or more positive) electron gain enthalpy.

Consider the following changes:
The second ionization energy of M could be calculated from the energy values associated with:
  • a)
    1 + 3 + 4
  • b)
    2 – 1 + 3
  • c)
    1 + 5
  • d)
    5 – 3
Correct answer is option 'D'. Can you explain this answer?

Dronacharya Institute answered
M (s) → M (g)     ........(1)
M (s) → M2+ (g) + 2e-   .......(2)
M (g) → M+ (g) + 2e-   .......(3)
M+ (g) → M2+ (g) + e-  .......(4)
M (g) → M2+ (g) + 2e- .......(5)

As we are interested in 2nd I.E.
i.e. Energy required to form M2+ (g) from M+ (g)
∴ 4 is correct.
but this option doesn't exist.
⇒ 1 + 3 + 4 is also wrong
1 + 5 gives total energy for M(g) → M2+ (g)
∴ it is also wrong.
3rd reaction is wrong,
∴ 2 - 1 + 3 also dismised.
⇒ (d) is correct

Calculate the among of energy required to convert 110 mg of ‘X’ atom in gaseous state into X+ ion.
  • a)
    10.4 kJ
  • b)
    12.3 kJ
  • c)
    11.3 kJ
  • d)
    14.5 kJ
Correct answer is option 'C'. Can you explain this answer?

Ameya Reddy answered
Explanation:
The melting point of a compound is dependent on several factors such as the strength of the intermolecular forces, size and charge of the ions, and lattice energy. In this case, the melting point of NaF is higher than the other three compounds due to the following reasons:

Ionic Character:
NaF has the highest ionic character among the given compounds. This is because the difference in electronegativity between Na and F is the highest among the given compounds. As a result, the bond between Na and F is highly polar covalent, and the F ion has a high negative charge. This leads to stronger electrostatic forces of attraction between the oppositely charged ions, resulting in a higher melting point.

Lattice Energy:
The lattice energy of NaF is the highest among the given compounds. This is because the F ion is the smallest among the given anions, and NaF has the highest charge density. The high charge density leads to a higher attraction between the ions, resulting in a higher lattice energy. The high lattice energy contributes to the high melting point of NaF.

Size and Charge of Ions:
The size and charge of the ions also play a role in the melting point of a compound. In this case, NaBr and NaI have larger anions, which means that the distance between the ions in the lattice is greater. This leads to weaker electrostatic forces of attraction and a lower melting point. Additionally, NaCl and NaBr have smaller charges on their anions, resulting in weaker electrostatic forces and lower lattice energies.

In conclusion, NaF has the highest melting point among the given compounds due to its high ionic character, high lattice energy, and small size of the F ion.

When magnesium burns, in air, compounds of magnesium formed are magnesium oxide and:
  • a)
    Mg3N2
  • b)
    MgCO3
  • c)
    Mg(NO3)2
  • d)
    MgSO4
Correct answer is 'A'. Can you explain this answer?

Rahul Chatterjee answered
There’s only one product, I guess.

2Mg + O2 → 2MgO

This is, of course, considering pure oxygen.

If magnesium is burnt in air (which contains 78% N2), a side-product Mg3N2

The correct order of second I. E. of C, N, O and F are in the order:
  • a)
    F > O > N > C
  • b)
    C > N > O > F
  • c)
    O > N > F > C
  • d)
    O > F > N > C
Correct answer is option 'D'. Can you explain this answer?

Aaron B Mathew answered
The electronic configuration of oxygen is 1s2 2s2 2p4 on removing one electron we get a half filled configuration which is stable. So among these it will b the hardest to remove the second electron of oxygen. The size of fluorine is less than nitrogen. On removing one electron, its size will decrease again and so, its second I.E will be greater than nitrogen. carbon has the least second I.E because on removing the second electron, it will complete octet and attain stability

Which one of the following elements shows both positive and negative oxidation states:
  • a)
    Cesium
  • b)
    Fluorine
  • c)
    Iodine
  • d)
    Xenon
Correct answer is option 'C'. Can you explain this answer?

Sagarika Patel answered
All the other halogens-chlorine, bromine and iodine- exhibit positive oxidation states upto +7, apart from the most stable negative state of -1. But fluorine is more electronegative than any other element in the periodic table and so it cannot be assigned a positive oxidation number in its compounds.

The size of isoelectronic species: F–, Ne and Na+ is affected by:
  • a)
    Nuclear charge (n)
  • b)
    Valence principal quantum number(n)
  • c)
    Electron-electron interaction in the outer orbitals
  • d)
    None of the factor because their size is the same 
Correct answer is option 'A'. Can you explain this answer?

Shivam Sharma answered
nuclear charge (Z). Isoelectronic species are the species belonging to different atoms or ions which have same number of electrons but different magnitudes of nuclear charges.
The size of an isoelectronic species increases with a decrease in the nuclear charge (Z). For example, the order of the increasing nuclear charge of F–, Ne, and Na+ is as follows:

       F– < Ne < Na+

Z     9     10     11

Therefore, the order of the increasing size of F–, Ne and Na+ is as follows:

Na+ < Ne < F–

The most electropositive element possesses the electronic configuration:
  • a)
    [He]2s1
  • b)
    [Ne]3s2
  • c)
    [Xe]6s1
  • d)
    [Xe]6s2
Correct answer is option 'C'. Can you explain this answer?

Mrinalini Sen answered
The electronic configuration [Xe]6s^1 is the configuration of Cs which is highly electropositive element of alkali metal family. It has the lowest ionisation potential in its family.

The second electron gain enthalpies (in kJ mol–1) of oxygen and sulphur respectively are:
  • a)
    –780, + 590
  • b)
    –590, + 780
  • c)
    +590, + 780
  • d)
    +780, +590
Correct answer is option 'D'. Can you explain this answer?

Nandini Das answered
The second electron gain enthalpy (electron affinity) is the energy released when a neutral atom gains an electron to form a negative ion with a 1- charge.

Explanation:
- Oxygen has 6 electrons in its valence shell and needs 2 more electrons to complete its octet.
- When it gains the first electron, it becomes a negatively charged ion, O-.
- However, when it gains the second electron, it has to overcome a repulsive force from the already existing negative charge. This makes it more difficult for oxygen to gain the second electron, hence it has a higher second electron gain enthalpy value.
- Sulphur, on the other hand, has 6 electrons in its valence shell and needs 2 more electrons to complete its octet.
- When it gains the first electron, it becomes a negatively charged ion, S-.
- When it gains the second electron, it is easier for sulphur to gain it as compared to oxygen, hence it has a lower second electron gain enthalpy value.

Therefore, the correct answer is option 'D' where oxygen has a higher second electron gain enthalpy of 780 kJ mol-1 and sulphur has a lower second electron gain enthalpy of 590 kJ mol-1.

Which of the following statements is/are wrong:
  • a)
    Van der Waals radius of iodine is more than its covalent radius
  • b)
    All isoelectronic ions belong to same period of the periodic table
  • c)
    I.E1 of N is higher than that of O while I.E2 of O is higher than that of N
  • d)
    The electron affinity N is almost zero while that of P is 74.3 kJ mol–1
Correct answer is option 'D'. Can you explain this answer?

Aditya Deshmukh answered
Well, noble gases such as Helium, Neon and Argon have an electron affinity nearly zero because they have octet configuration. Due to this reason , they find it hard to attaract electrons. However higher members of Group 18 in the periodic table such Xenon, krypton and radon, do form compound like XeO2 and XeO4 due to the presence of vacant d- orbitals.

Consider the following information about element P and Q:
Then formula of the compound formed by P and Q element is:
  • a)
    PQ
  • b)
    P3Q2
  • c)
    P2Q3
  • d)
    PQ2
Correct answer is option 'C'. Can you explain this answer?

Surendra Bind answered
Group 15write the configuration in outer shell in 5clectron so element in vacancy 3&5but 3valancy is more stable because absence is d orbital (in case of Nitrogen) and group 2write the configuration in outer shell in 2 electron so this vacancy is 2 so the formula of the compound is formula P2Q3

Which of the following species must have maximum number of electrons in ‘dxy’ orbital:
  • a)
    Cr
  • b)
    Fe3+
  • c)
    Cu+
  • d)
    Both a and b
Correct answer is option 'C'. Can you explain this answer?

Sinjini Singh answered
Explanation:
The maximum number of electrons in the dxy orbital is 2. To determine which of the given species has the maximum number of electrons in the dxy orbital, we need to look at their electronic configurations.


Electronic Configurations:
a) Cr: [Ar] 3d5 4s1

b) Fe3+: [Ar] 3d5

c) Cu: [Ar] 3d10 4s1


From the electronic configurations, we can see that:


  • Cromium (Cr) has 5 electrons in the d orbital (d1, d2, d3, d4, d5)

  • Iron (Fe3+) has 5 electrons in the d orbital (d1, d2, d3, d4, d5)

  • Copper (Cu) has 10 electrons in the d orbital (d1, d2, d3, d4, d5, d6, d7, d8, d9, d10)



Therefore, the species with the maximum number of electrons in the dxy orbital is Copper (Cu) because the dxy orbital is one of the five d orbitals, and copper has a completely filled d orbital with 10 electrons.


Hence, the correct answer is option 'C'.

Consider the following electronic configuration of an element (P):
[Xe] 4f145d16s2
The correct statement about element ‘P’ is:
  • a)
    It belongs to 6th period and 1st group
  • b)
    It belongs to 6th period and 2nd group
  • c)
    It belongs to 6th period and 3rd group
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Anirban Kapoor answered
Electronic Configuration of Element P

The electronic configuration of element P is: [Xe] 4f14 5d1 6s2.

Determining the Period and Group of Element P

To determine the period and group of element P, we need to look at its valence shell configuration, which is 6s2.

Period: The valence shell of element P is in the sixth energy level, so it belongs to the sixth period.

Group: The valence shell of element P has two electrons in the s orbital, so it belongs to the third group (also known as group 13 or the boron group).

Therefore, the correct statement is option C: Element P belongs to the 6th period and 3rd group.

Which electronic configuration must represent an atom in an excited state:
  • a)
    1s2, 2s2 2p1
  • b)
    1s2, 2s2 2p2
  • c)
    1s2, 2s2 2p2, 3s1
  • d)
    1s2, 2s2 2p5
Correct answer is option 'C'. Can you explain this answer?

Sahil Kapoor answered
Excited State Electronic Configuration

An atom in an excited state has at least one electron in a higher energy level than the ground state electronic configuration.

Explanation of Options

a) 1s2, 2s2 2p1 - This is the ground state electronic configuration of carbon. It has all its electrons in the lowest possible energy levels.

b) 1s2, 2s2 2p2 - This is also the ground state electronic configuration of oxygen.

c) 1s2, 2s2 2p2, 3s1 - This represents an excited state of aluminum. One of the electrons from the 3s orbital has been promoted to the 3p orbital.

d) 1s2, 2s2 2p5 - This is the ground state electronic configuration of fluorine.

Conclusion

The correct answer is option 'C' because it represents an atom in an excited state.

Sodium generally does not shown oxidation state of +2, because of its:
  • a)
    High first ionization potential
  • b)
    High second ionization potential
  • c)
    Large ionic radius
  • d)
    High electronegativity
Correct answer is option 'B'. Can you explain this answer?

Jughs answered
Once sodium loses 1 electron..it reaches its nearby noble gas configuration (ie. of Ne)..now becomes stable...won't prefer to lose an electron..hence high 2nd IP.

First the ionization energies (in kJ/mol) of three representative elements are given below:
Then incorrect option is:
  • a)
    Q: Alkaline earth metal
  • b)
    P: Alkali metals
  • c)
    R: s-block element
  • d)
    They belong to same period
Correct answer is option 'C'. Can you explain this answer?

Sanjeev Kumar Yadav answered
In P difference between 1st and 2nd ionisation enthalpy is high it means it is difficult to remove 2nd electron from P it means alkali metal

similarly Q is alkaline earth metal.

similarly R is group 13 element and not sure block element

In which of the following arrangements, the order is not correct according to the property indicated against it:
  • a)
    Increasing size: Al3+ < Mg2+ < Na+ < F–
  • b)
    Increasing I.E.1: B < C < N < O
  • c)
    Increasing E.A1: I < Br < F < Cl
  • d)
    Increasing metallic radius: Li < Na < K < Rb
Correct answer is option 'C'. Can you explain this answer?

Ameya Reddy answered
A) Increasing size: Al3+ < mg2+="" />< />

This arrangement is not correct according to the property of increasing size. The correct arrangement should be Na+ < mg2+="" />< al3+="" as="" the="" size="" of="" the="" ion="" increases="" with="" the="" addition="" of="" more="" electrons="" and="" protons.="" na+="" has="" the="" smallest="" size,="" followed="" by="" mg2+="" and="" then="" al3+.="" al3+="" as="" the="" size="" of="" the="" ion="" increases="" with="" the="" addition="" of="" more="" electrons="" and="" protons.="" na+="" has="" the="" smallest="" size,="" followed="" by="" mg2+="" and="" then="" />

The period number and group number of “Tantalum” (Z = 73) are respectively:
  • a)
    5, 7
  • b)
    6, 13
  • c)
    6, 5
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Anisha Banerjee answered
Period and Group of Tantalum

Periodic Table of Elements

The periodic table of elements is a tabular arrangement of chemical elements based on the periodic trends in their properties. The elements are arranged in rows and columns according to their atomic number, electron configuration, and chemical properties.

Tantalum (Ta)

Tantalum is a chemical element with the symbol Ta and atomic number 73. It is a rare, hard, blue-gray, lustrous transition metal that is highly corrosion-resistant. Tantalum is used in the production of capacitors, surgical implants, and nuclear reactors.

Period Number of Tantalum

The period number of an element in the periodic table is the number of electron shells it has. Tantalum has an atomic number of 73, which means it has 73 electrons arranged in different shells. The period number of an element is equal to the number of shells it has. Therefore, the period number of tantalum is 6.

Group Number of Tantalum

The group number of an element in the periodic table is the number of valence electrons it has. Valence electrons are the electrons in the outermost shell of an atom that are involved in chemical bonding. Tantalum belongs to group 5 of the periodic table because it has 5 valence electrons. Therefore, the group number of tantalum is 5.

Conclusion

The correct answer is option C, which states that the period number and group number of tantalum are 6 and 5, respectively.

Aqueous solutions of two compounds M1—O—H and M2—O—H are prepared in two different breakers. If the electronegativity of M1 = 3.4, M2 = 1.2, O = 3.5, and H = 2.1, then the nature of two solutions will be respectively:
  • a)
    Acidic, Basic
  • b)
    Acidic, Acidic
  • c)
    Basic, Acidic
  • d)
    Basic, Basic
Correct answer is option 'A'. Can you explain this answer?

Anirban Kapoor answered
Explanation:
Electronegativity is the measure of the tendency of an atom to attract a bonding pair of electrons. The higher the electronegativity of an atom, the more it attracts electrons towards itself.

The electronegativity of M1 and M2 are 3.4 and 1.2 respectively. Since M1 has a higher electronegativity, it will have a stronger attraction towards the O-H bond and will be able to pull the electron density towards itself. This will result in the release of H+ ions and make the solution acidic.

On the other hand, M2 has a lower electronegativity and will have a weaker attraction towards the O-H bond. As a result, it will not be able to release H+ ions and the solution will remain basic.

Therefore, the solution of M1OH will be acidic and the solution of M2OH will be basic.

Summary:
- M1OH solution will be acidic
- M2OH solution will be basic

In the fourth period of the periodic table, how many elements have one or more 4d electrons:
  • a)
    2
  • b)
    18
  • c)
    0
  • d)
    6
Correct answer is option 'C'. Can you explain this answer?

Harshitha Sharma answered
Answer:

In the fourth period of the periodic table, there are no elements that have one or more 4d electrons. This is because the fourth period corresponds to the n=4 energy level, which includes the 4s and 3d sublevels.

Explanation:

- The fourth period of the periodic table includes the elements potassium (K) through krypton (Kr).
- The 4s sublevel can hold up to 2 electrons, and the 3d sublevel can hold up to 10 electrons.
- Therefore, the elements in the fourth period can have up to 12 electrons in the n=4 energy level.
- However, none of the elements in the fourth period have any electrons in the 4d sublevel.
- The 4d sublevel comes after the 3d sublevel in energy, so it is not filled until the fifth period.
- Elements in the fifth period, such as niobium (Nb) and ruthenium (Ru), have electrons in the 4d sublevel.
- Therefore, the correct answer is option 'C', which states that there are no elements in the fourth period with 4d electrons.

Which of the following represents the correct order of increasing first ionization enthalpy for Ca,Ba,S,Se andAr?
  • a)
    Ca<S<Ba<Se<Ar
  • b)
    S<Se<Ca<Ba<Ar
  • c)
    Ba<Ca<Se<S<Ar
  • d)
    Ca<Ba<S<se<Ar
Correct answer is option 'C'. Can you explain this answer?

Jaya Sen answered
The decreasing order of the first ionization energy of the given elements is Xe, As, Be, Al. This can be explained by the following points:

1. Nuclear Charge: The first ionization energy of an element depends on the nuclear charge of the atom. The higher the nuclear charge, the more difficult it is to remove an electron from the atom. Xe has the highest nuclear charge among the given elements, followed by As, Be, and Al.

2. Electron Shielding: The first ionization energy also depends on the number of electrons in the outermost shell of the atom and the number of inner shells. The outer electrons are shielded from the nucleus by the inner electrons. The more the number of inner shells, the more the shielding effect. Xe has a larger number of inner shells, resulting in a stronger shielding effect, making it easier to remove an electron from the outermost shell.

3. Atomic Radius: The first ionization energy is inversely proportional to the atomic radius of the element. The larger the atomic radius, the easier it is to remove an electron from the outermost shell. As and Al have smaller atomic radii than Xe and Be, making it harder to remove an electron from their outermost shells.

Based on the above factors, the decreasing order of the first ionization energy of the given elements is Xe, As, Be, Al. Therefore, option C (Xe, As, Be, Al) is the correct answer.

The fourth period ends with the element ________
  • a)
    Helium
  • b)
    Argon
  • c)
    Neon
  • d)
    Krypton
Correct answer is option 'D'. Can you explain this answer?

Mihir Singh answered
The fourth period is the row of elements in the periodic table that starts with potassium (K) and ends with krypton (Kr). Therefore, the correct answer is option 'D' which is krypton.

Explanation:

Periods in the periodic table are the horizontal rows that are numbered from 1 to 7. The fourth period is the row that follows the third period and is located between the third and fifth periods. This period contains 18 elements starting from potassium (K) and ending with krypton (Kr).

Krypton is the last element in the fourth period, which means that all of the elements that come after krypton belong to the fifth period. Krypton is a noble gas that has an atomic number of 36 and is located in group 18 of the periodic table. It is a colorless, odorless, and tasteless gas that is relatively rare in the Earth's atmosphere.

It's important to note that the elements in a period have the same number of electron shells, which means that they have similar chemical properties. However, the number of protons and electrons in each element increases as you move from left to right in a period, which affects the physical and chemical properties of the elements.

In conclusion, krypton is the correct answer for this question as it is the last element in the fourth period and all elements beyond it belong to the fifth period.

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