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Test: Periodic Properties- 1


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20 Questions MCQ Test Inorganic Chemistry | Test: Periodic Properties- 1

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Test: Periodic Properties- 1 - Question 1

Which pair of elements has maximum electronegativity difference?

Detailed Solution for Test: Periodic Properties- 1 - Question 1
  • The general peroidic table electronegativity trend says that moving down the group electronegativity of element decreases.
  • So in the atoms Li , F , Na , Cl , Br the most electronegative is F and the least electronegative or most electropositive is Na . Among Li & Na, Na is more electropositive. Among F, Cl & Br, F is most electronegative. 
  • So, Na & F will have maximum electronegativity difference. 

Test: Periodic Properties- 1 - Question 2

When magnesium burns, in the air, compounds of magnesium formed are magnesium oxide and ______.

Detailed Solution for Test: Periodic Properties- 1 - Question 2

When magnesium burns in the air it forms magnesium nitride with nitrogen and with oxygen it forms magnesium oxide. 

  • The reaction of magnesium with oxygen to form magnesium oxide is as follows:
    2Mg+O2→2MgO
  • The reaction of magnesium with nitrogen in air to form magnesium nitride is as follows:

    3Mg+N2→Mg3N2

Test: Periodic Properties- 1 - Question 3

The set representing the correct order of ionic radius is:

Detailed Solution for Test: Periodic Properties- 1 - Question 3
  • In general, the ionic radius increases on moving from top to bottom in the group. 
    So, radius of Na+ > Li+
    and radius of Mg2+ > Be2+
  • In general, the ionic radius decreases on moving from left to right in a period. 
  • Also, greater the positive charge, lesser the radius. Hence, 
    Na+ > Li+ > Mg2+ > Be2+

The ionic radius of the following elements are:
Na+: 1.02 Ao
Be2+: 0.39 Ao
Mg2+: 0.72 Ao
Li+: 0.76 Ao
So, the correct option is B.

Test: Periodic Properties- 1 - Question 4

Which of the following pairs has the highest difference in their first ionization energy?

Detailed Solution for Test: Periodic Properties- 1 - Question 4
  • I.E. increases from left to right and decreases from top to bottom in the periodic table. 
  • Ne (noble gas) and Na (alkali metal) belong to 2nd and 3rd period respectively and show highest difference in their first I.E.
  • Subsequent members of the same group show lower I.E. on moving down in the group
Test: Periodic Properties- 1 - Question 5

True increasing order of acidity of the oxides of Mn is:

Detailed Solution for Test: Periodic Properties- 1 - Question 5
  • The correct increasing order of acidity of Mn oxide is MnO < MnO2 < Mn2O7
  • With increasing (+)ve oxidation state, acidity increases.
  • Mn in  Mn2O7 the oxidation state is +7 state, Mn in MnO the oxidation state is +2, The oxidation number of 
    Mn in MnO2 is +4.
Test: Periodic Properties- 1 - Question 6

The correct order of ionic size of N3–, Na+, F, Mg2+ and O2– is:

Detailed Solution for Test: Periodic Properties- 1 - Question 6
  • If the ions derived from different atoms are isoelectronic species, then they all have the same number of electrons in their electronic shells and will have got the same electronic configuration. 
  • But their nuclear charge will differ because of their difference in the number of protons in the nucleus. 
  • With an increase in the number of protons in the nucleus, the electrons are more attracted towards the 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.

Test: Periodic Properties- 1 - Question 7

The size of isoelectronic species, F, Ne and Na+ is affected by:

Detailed Solution for Test: Periodic Properties- 1 - Question 7
  • Isoelectronic species are the species belonging to different atoms or ions which have the 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).
    Example:
     The order of the increasing nuclear charge of F, Ne, and Na+ is as follows: 
    F < Ne < Na+
    where Z = 9, 10, 11 respectively.
  • Therefore, the order of the increasing size of F, Ne and Nais as follows:
    Na+ < Ne < F
Test: Periodic Properties- 1 - Question 8

The first ionization potential in electron volts of nitrogen and oxygen atoms are respectively given by:

Detailed Solution for Test: Periodic Properties- 1 - Question 8

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

Test: Periodic Properties- 1 - Question 9

Which of the following ions is most unlikely to exist?

Detailed Solution for Test: Periodic Properties- 1 - Question 9

Be- ion is most unlikely to exist as Be has a higher tendency to lose an electron and Be has a full filled configuration.

Li- =  1s2, 2s2 (In it all sub shells are saturated so, it is stable)

Be- =  1s2, 2s2 2p1

B-  =  1s2, 2s2 2p2

C-   =  1s2, 2s2 2p3

Test: Periodic Properties- 1 - Question 10

 Which of these does not reflect the periodicity of the elements?

Detailed Solution for Test: Periodic Properties- 1 - Question 10

The neutron/proton ratio does not reflect the periodicity of the elements. 

Test: Periodic Properties- 1 - Question 11

What is the atomic number of the element with the maximum number of unpaired 4p electron?

Detailed Solution for Test: Periodic Properties- 1 - Question 11

The electronic configuration is given as follows:

  • Electronic Configuration of element having Atomic number 33: 1s2 2s2p3s3p6 4s2 3d10 4p3
  • Electronic Configuration of element having Atomic number 26: 1s2s2 2p6 3s2 3p4s2 3d6
  • Electronic Configuration of element having Atomic number 23:  1s2s2 2p6 3s2 3p4s2  3d3
  • Electronic Configuration of element having Atomic number 15: 1s2 2s2 2p6 3s2 3p3

Observation of the above given 4 E.C. tells us that the element having atomic number 33 will have the maximum number of unpaired electrons i.e. 3.

Test: Periodic Properties- 1 - Question 12

The electronic configuration of four elements are:
(I) [Kr]5s1
(II) [Rn]5f146d17s2
(III) [Ar]3d104s24p5
(IV) [Ar]3d64s2
Consider the following statements:
(i) I shows variable oxidation state.
(ii) II is a d-block element.
(iii) The compound formed between I and III is covalent.
(iv) IV shows single oxidation state.

Which statement is True(T) or False (F)?

Detailed Solution for Test: Periodic Properties- 1 - Question 12

Explanation: (I) [Kr]5s1, shows only single oxidation state +1.
(II) [Rn]5f14 6d1 7s2, it is f-block element (Z=103).
(III) The compound formed between I and Ill is ionic.
(IV) [Ar]3d6 4s2, (Z=26) Fe shows variable oxidation state.
Therefore, all the above statements in the question are false.

Test: Periodic Properties- 1 - Question 13

If period number and group number of any representative element(s) are same then which of the following statement is incorrect regarding such type element(s) in their ground state: (Period number and group number are according to modern form of periodic table)

Detailed Solution for Test: Periodic Properties- 1 - Question 13

When the period and group is the same, it can only be for Hydrogen(1s1) and Beryllium(1s2 2s2). Then: 
Principal quantum number (n) = 2,
Azimuthal quantum number (l) = 0,
Magnetic quantum number (m) = 0,
Spin quantum number (s) = +1 / 2 or -1 / 2.

Test: Periodic Properties- 1 - Question 14

How does the energy gap between successive energy levels in an atom vary from low to high values?

Detailed Solution for Test: Periodic Properties- 1 - Question 14
  • 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 energyso the smaller it is, the smaller the difference with neighbouring levels will be.
Test: Periodic Properties- 1 - Question 15

Which of the following properties of the alkaline earth metals increase from Be to Ba:
(I) Atomic radius
(II) Ionisation energy
(III) Nuclear charge

Detailed Solution for Test: Periodic Properties- 1 - Question 15
  • Effective nuclear charge increases down the group and atomic radii increases due to the addition of a new subshell.
  • Whereas on moving down the group, the force of attraction between the electrons and the nucleus decreases. Hence ionisation energy decreases.
Test: Periodic Properties- 1 - Question 16

In which of the following pair, both the species are isoelectronic but the first one is large in size than the second?

Detailed Solution for Test: Periodic Properties- 1 - Question 16
  • F- and Na+ both have 10 electrons. Atomic no. of F is 9 and for Na it is 11.
  • So for F-, 9 protons are pulling 10 electrons and for Na+,11 proton are pulling 10 electrons. So attractive force is more for Na+ compared to F-.
  • So the radius of F- is more than Na+.
Test: Periodic Properties- 1 - Question 17

The order of increasing ionic radius of the following is:

Detailed Solution for Test: Periodic Properties- 1 - Question 17
  • For Mg2+and Al3+ you can just check (z/e) ratio as these two are isoelectronic species, and you must know that Li+ is slightly greater than Mg2+, this is experimental data, the reason is out of your syllabi as it includes Schrodinger's eqⁿ.
  • Well K+ has the configuration of Ar[18] which makes it larger among them as noble gases are abnormally larger than their neighbour elements due to extra stability.
Test: Periodic Properties- 1 - Question 18

If the ionic radii of K+ and F are nearly the same (i.,e. 1.34 Å), then the atomic radii of K and F respectively are:

Detailed Solution for Test: Periodic Properties- 1 - Question 18
  • If any atom loss electron then its radius will be decreased so, the radius of atom K has more value than the K+ ion.
  • If any atom gain electron then its size increases, it means the radius of the F atom is smaller than the F- ion.
Test: Periodic Properties- 1 - Question 19

Among the following species, Identify the isoelectronic and isostructural pair.

Detailed Solution for Test: Periodic Properties- 1 - Question 19

The correct option is B [BF3, NO-3]
Solution: 
Isoelectronic species have same number of electrons and Isostructural species have same hybridisation at the central atom.
Steric Number (x) = 1/2(M + V − C+A).
where, M = number of monovalent surrounding atoms
V = number of valence electrons on the central atom
C = charge (with sign)
A=charge on anion

NF3
number of electron=7+9×3=34
Steric Number (x)=1/2(3+5−0+0).
Steric Number (x)=4.
Hybridisation =sp3

NO-3
number of electron=7+8×3+1=32
Steric Number (x)=1/2(0+5−0+1).
Steric Number (x)=3.
Hybridisation =sp2

BF3
number of electron=5+9×3=32
Steric Number (x)=1/2(3+3−0+0).
Steric Number (x)=3.
Hybridisation =sp2

[BF3, NO-3] are isoelectronic and isostructural pairs.

Test: Periodic Properties- 1 - Question 20

The correct set of decreasing order of electronegativity is:

Detailed Solution for Test: Periodic Properties- 1 - Question 20

The correct option is C)

  • The correct set of decreasing order of electronegativity is H>Li>Na
  • In a group, electronegativity decreases from top to bottom due to increase in the atomic size.
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