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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

An element belongs to period 2 and group 2, then the number of valence electrons in the atoms of this element is:        

Detailed Solution for Test: Periodic Properties- 1 - Question 1
  • The element belongs to group 2 and period 2 is Be( atomic number 4).
  • Thus 2 electrons will be accommodated in the first and the other 2 in the next shell(that is, it is the outer or valence shell). The valence shell contains 2 electrons.
Test: Periodic Properties- 1 - Question 2

Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following does not affect the valence shell?

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

Nuclear mass does not affect the valence electrons because the nuclear mass is so small, it is considered as negligible. 

Test: Periodic Properties- 1 - Question 3

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

Detailed Solution for Test: Periodic Properties- 1 - Question 3

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 4

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

Detailed Solution for Test: Periodic Properties- 1 - Question 4
  • 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 5

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

Detailed Solution for Test: Periodic Properties- 1 - Question 5

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

Test: Periodic Properties- 1 - Question 6

Which of the following ions is most unlikely to exist?

Detailed Solution for Test: Periodic Properties- 1 - Question 6

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 7

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 that 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.

Detailed Solution for Test: Periodic Properties- 1 - Question 7
  • The electronegativities decrease from X to Y to Z.
  • X, Y and Z could be phosphorus aluminium and sodium respectively.
Test: Periodic Properties- 1 - Question 8

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

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

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

Test: Periodic Properties- 1 - Question 9

If the Aufbau principle had not been followed, Ca (Z = 20) would have been placed in the:

Detailed Solution for Test: Periodic Properties- 1 - Question 9
  • Electronic configuration of Ca is:
    1s22s22p63s23p64s2
  • The 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 1s22s22p63s23p63d and it belongs to d block.
Test: Periodic Properties- 1 - Question 10

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

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

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 11

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 11

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 12

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 12

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 13

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 13
  • 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 14

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 14
  • 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 15

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

Detailed Solution for Test: Periodic Properties- 1 - Question 15

With increasing (+)ve oxidation state, acidity increases.

Test: Periodic Properties- 1 - Question 16

The set representing the correct order of ionic radius is:

Detailed Solution for Test: Periodic Properties- 1 - Question 16

In general, the ionic radius increases on moving from top to bottom in the group and decreases on moving from left to right in a period. 
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 17

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 17
  • 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 18

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

Detailed Solution for Test: Periodic Properties- 1 - Question 18
  • 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 19

The order of increasing ionic radius of the following is:

Detailed Solution for Test: Periodic Properties- 1 - Question 19
  • 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 20

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 20
  • 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.
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