Important Questions for CSIR NET Chemical Science 2022

Most Important Questions on Inorganic Chemistry - Download PDF

1. Which of the following order is correct?

Mg²⁺(size) < Al³⁺ (size)
S (E.A.) < O (E.A.)
Hg (I.E) > Cd (I.E)
S (I.E) > P (I.E)

2. The correct order of acidic strength is

XeF₆ >XeO₂F₄ >XeO₃F₂ >XeO₄
XeF₆ >XeO₃F₂ XeO₂F₄ >XeO₄
XeO₄ >XeO₃F₂ >XeO₂F₄ >XeF₆
XeO₄ >XeO₃F₂ >XeF₆ >XeO₂F₄

3. The X formed in the given reaction is

NaHPO₄
NaP₂O₇
Na₄P₂O₇
NaHPO₂

4. Which of the following compounds will form silicone having the highest molecular weight?

RSi(OH)₃
R2Si(OH)₂
R₃Si(OH)
R₄Si

5. On reaction with NH₃, NH₄NO₃ and KNH₂ will act as

Acid and acid
Acid and base
Base and acid
Base and base

6. Which of the following is not correct about cytochrome P-450?

Facilitates the cleavage of O₂.
Absorbs at 450 nm with their CO complexes
It contains a histidine ligand.
Converts insoluble hydrocarbons to water-soluble alcohols.

7. The order of stability of complexes of porphyrins with +2 metal ions is:

Zn²⁺< Cu²⁺> Ni²⁺> Co²⁺>Fe2+
Cu2+< Zn2+> Ni2+> Co2+>Fe2+
Ni2+> Cu2+> Co2+> Fe2+>Zn2+
Ni2+< Zn2+> Cu2+> Co2+>Fe2+

8. The term symbol for the ground state of Ce3+ is

F3/2
2F3/2
2F5/2
F5/2

9. Determine the geometry of the complex [HOs5(CO)15]-.

Trigonal Prism
Trigonal Bipyramidal
Square pyramidal
Pentagonal bipyramidal

10. The most suitable route to prepare cis isomer of [Pt(C₂H₄)(NH₃)Cl₂]

[PtCl₄]^2- with C2H4 followed by reaction with NH₃
[PtCl₄]^2- with NH₃ followed by reaction with C₂H₄
[Pt(NH₃)₄]²⁺ with Cl^- followed by reaction with C₂H₄
[Pt(NH₃)₄]²⁺ with C₂H₄ followed by reaction with Cl

Solutions:

Solution 1.

A. More the positive charge on cation smaller is the size of cation. Thus, the size of Mg2+ > Al3+. Hence, this is incorrect.

O has more electronegativity than S. However, O is smaller than S, hence the charge density is more on anion of O than S. Because of more charge density, repulsion among the electrons increases in anion of O. However, charge density in anion of S is less. Hence this is incorrect.
Ionisation energy of Hg is more than Cd, because in Hg due to the presence of 4f it has poor shielding effect, which brings higher effective nuclear charge on the outermost valence shell in Hg. To pull electron from outer shell of Hg, more energy needs to be supplied. Hence this is true.
P has three unpaired electrons in 3 p orbital that means it has stable half-filled electronic configuration, whereas sulphur has four electrons in 3 p orbital i.e., does not have half-filled configuration. Therefore, it is easier to remove an electron from a sulphur atom as compared to phosphorus. Hence, the ionisation energy of S is less than P. Hence, this option is incorrect.

Solution 2.

The acidic strength of xenon fluorides and xenon oxyfluorides depends on the number of F atoms directly attached to Xe. As F is highly electronegative, it reduces the electron density on the central Xe atom and makes it more acidic.
XeF6 >XeO2F4 >XeO3F2 >XeO4
Thus, the correct answer is (A).

Solution 3.

Solution 4.

RSi(OH)3: It will result in the formation of branched polymers with removal of water molecules. Thus, this silicone will have the highest molecular weight.
R2Si(OH)2: It will form a linear chain polymer.
R3Si(OH): It will only form dimeric compounds.
It is tetra alkyl silane, and exists in monomeric form.

Solution 5.

Therefore, the correct answer is B.

Solution 6.

Statement C is incorrect because it contains cystine ligand and not histidine.

The structure of cytochrome P-450 is given below:

Solution 7. The order of stability of complexes is in accordance with the Irving William Series

Fe2+ < Co2+< Ni2+<Cu2+> Zn2+

But here Ni2+ is most stable because of the formation of a square planar complex followed by the Cu2+, Co2+& Fe2+ and Zn is least stable because it has filled d-orbital.

Solution 8. The term symbol is an abbreviated description of the total angular quantum numbers in a multi-electron atom. Each of the arrangements obtained by combining the resultant L and S terms. Corresponds to an electronic arrangement sometimes called a spectroscopic state which is expressed by the full-term symbol.

Electronic configuration of Ce3+=[Xe] 4f1

f-subshell has 7orbitals have

For f-orbital; L=3 +3 +2 +1 0 -1 -2 -3

Spin of single electron is S= ½

Spin multiplicity (s) =2S+1= 2(1/2) +1=2 →S=2

J= L-S (because orbitals are not half-filled)

J= 3-1/2= 5/2

Therefore, ground state term symbol for Ce3+ is= 2F5/2

Solution 9. Determination of structure for Geometry: -

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

E.g., H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step II → Compare with table of geometries.

[HOs5(CO)15]-

Electronic configuration of Os = 5d66s2

Hence, electronic configuration by Os= 8 e-

TVE = (8 X 5) + 1 + (15 X 2) + 1

= 40 + 1 + 30 +1

= 72

Compare TVE with table as:

It is Trigonal Bipyramidal in shape.

Solution 10. The preparation of cis isomer of [Pt(C2H4)(NH3)Cl2] is based on the trans directing ability of the ligand present. The order of trans-directing ability is C2H4> Cl-> NH3

cis - [Pt(C2H4)(NH3)Cl2] is formed by heating [PtCl4]2-with NH3 followed by C2H4