Chemical Bonding and Molecular Structure MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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321.
Which of the following molecules shows intramolecular hydrogen bonding?
A
$$o$$ - Nitrophenol
B
$$p$$ - Nitrophenol
C
Benzoic acid
D
Ethanol
Answer :
$$o$$ - Nitrophenol
322.
The $$BC{l_3}$$ is a planar molecule whereas $$NC{l_3}$$ is pyramidal, because
A
$$B - Cl$$ bond is more polar than $$N - Cl$$ bond
B
$$N - Cl$$ bond is more covalent than $$B - Cl$$ bond
C
nitrogen atom is smaller than boron atoms
D
$$BC{l_3}$$ has no lone pair but $$NC{l_3}$$ has a lone pair of electrons
Answer :
$$BC{l_3}$$ has no lone pair but $$NC{l_3}$$ has a lone pair of electrons
No lone pair of electrons is available in $$BC{l_3}.$$
One lone pair of electrons is available on $$N$$ atom, it occupies a corner in the tetrahedral arrangement. Therefore, $$NC{l_3}$$ appears pyramidal in shape.
323.
The formation of molecular complex $$B{F_3} - N{H_3}$$ results in a change in hybridization of boron
A
from $$s{p^2}$$ to $$ds{p^2}$$
B
from $$s{p^2}$$ to $$s{p^3}$$
C
from $$s{p^3}$$ to $$s{p^2}$$
D
from $$s{p^3}$$ to $$s{p^3}d$$
Answer :
from $$s{p^2}$$ to $$s{p^3}$$
In $$B{F_3},B$$ is $$s{p^2}$$ hybridized with one empty $${p_z}$$ orbital. The empty $${p_z}$$ orbital of $$B{F_3}$$ can be filled by lone pair of molecules such as $$N{H_3}.$$ When this occurs a tetrahedral molecule or ion is formed which is $$s{p^3}$$ hybridized.
324.
In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed?
325.
Which of the following shapes of $$S{F_4}$$ is more stable and why?
A
(i), due to $$3\,lp{\text{ - }}bp$$ repulsions at $${90^ \circ }.$$
B
(ii), due to $$2\,lp{\text{ - }}bp$$ repulsions.
C
Both are equally stable due to $$2\,lp{\text{ - }}bp$$ repulsions.
D
Both are unstable since $$S{F_4}$$ has tetrahedral shape.
Answer :
(ii), due to $$2\,lp{\text{ - }}bp$$ repulsions.
In structure (ii), $$lp{\text{ - }}bp$$ repulsions are minimum.
326.
Which of the following are isoelectronic and isostructural?
$$NO_3^ - ,CO_3^{2 - },ClO_3^ - ,S{O_3}$$
A
$$NO_3^ - ,CO_3^{2 - }$$
B
$$S{O_3},NO_3^ - $$
C
$$ClO_3^ - ,CO_3^{2 - }$$
D
$$CO_3^{2 - },S{O_3}$$
Answer :
$$NO_3^ - ,CO_3^{2 - }$$
NOTE : Isoelectronic species have same number of electrons and isostructural species have same type of hybridisation at central atom.
$$NO_3^ - ;$$ No. of $${e^ - } = 7 + 8 \times 3 + 1 = 32,$$ hybridisation of $$N$$ in $$NO_3^ - $$ is $$s{p^3}$$
$$CO_3^{2 - };$$ No. of $${e^ - } = 6 + 8 \times 3 + 2 = 32,$$ hybridisation of $$C$$ in $$CO_3^{2 - }$$ is $$s{p^3}$$
$$ClO_3^ - ;$$ No. of $${e^ - } = 17 + 8 \times 3 + 1 = 42,$$ hybridisation of $$Cl$$ in $$ClO_3^ - $$ is $$s{p^3}$$
$$S{O_3};$$ No. of $${e^ - } = 16 + 8 \times 3 = 40,$$ hybridisation of $$S$$ in $$S{O_3}$$ is $$s{p^2}$$
$$\therefore NO_3^ - $$ and $$CO_3^{2 - }$$ are isostructural and isoelectronic.
327.
Which of the following does not show octahedral geometry?
A
$$S{F_6}$$
B
$$I{F_5}$$
C
$$SiF_6^{2 - }$$
D
$$S{F_4}$$
Answer :
$$S{F_4}$$
$$S{F_4}$$ has trigonal bipyramidal geometry. $$S{F_6},I{F_5}$$ and $$SiF_6^{2 - }$$ have octahedral geometry.
328.
The electronic configuration of four atoms are given in brackets :
$$\eqalign{
& P\left( {1{s^2}2{s^2}2{p^1}} \right);Q\left( {1{s^2}2{s^2}2{p^5}} \right) \cr
& R\left( {1{s^2}2{s^2}2{p^6}3{s^1}} \right);S\left( {1{s^2}2{s^2}2{p^2}} \right) \cr} $$
The element that would most readily form a diatomic molecule is
A
$$P$$
B
$$Q$$
C
$$R$$
D
$$S$$
Answer :
$$Q$$
By sharing of 1 electron each, both the atoms of $$Q$$ will get a complement octet.
329.
Arrange the following ions in the order of decreasing $$X – O$$ bond length, where $$X$$ is the central atom
A
$$ClO_4^ - ,SO_4^{2 - },PO_4^{3 - },SiO_4^ - $$
B
$$SiO_4^{4 - },PO_4^{3 - },SO_4^{2 - },ClO_4^ - $$
C
$$SiO_4^{4 - },PO_4^{3 - },ClO_4^ - ,SO_4^{2 - }$$
D
$$SiO_4^{4 - },SO_4^{2 - },PO_4^{3 - },ClO_4^ - $$