Chemical Bonding and Molecular Structure MCQ Questions & Answers in Inorganic Chemistry | Chemistry

The bond length of $$O - O$$  in $${H_2}{O_2}$$  is $$147.5\,pm,$$  in $${O_3}$$  is $$128\,pm$$  and in $${O_2}$$  it is $$121\,pm,$$  so the correct order is $${O_2} < {O_3} < {H_2}{O_2}.$$

In $$KCN,$$  ionic bond is present between $${K^ + }$$ and $$C{N^ - }$$ and covalent bonds are present between carbon and nitrogen $$C \equiv N$$ .

three angle below $$M$$  and three above $$M$$  hence $$= 6$$

$$Si{F_4}$$  and $$S{F_4}$$  are not isostructural because $$Si{F_4}$$  is tetrahedral due to $$s{p^3}$$  hybridisation of $$Si.$$
$${}_{14}Si = 1{s^2},2{s^2},2{p^6},3{s^2}3{p^2}$$      ( in ground state )
$${}_{14}Si = 1{s^2},2{s^2}2{p^6},3{s^1}3{p^3}$$      ( in excited state )

Hence, four equivalent $$s{p^3}$$  hybrid orbitals are obtained and they are overlapped by four $$p - {\text{orbitals}}$$   of four fluorine atoms on their axis. Thus, it shows following structure:

While $$S{F_4}$$  is not tetrahedral but it is arranged in trigonal bipyramidal geometry (has see saw shape) because in it $$S$$  is $$s{p^3}d$$  hybrid.
$${}_{16}S = 1{s^2},2{s^2}2{p^6},3{s^2}3p_x^23p_y^13p_z^1$$      ( in ground state )
       $$ = 1{s^2},2{s^2}2{p^6},\underbrace {3{s^2}3p_x^13p_y^13p_z^13d_{xy}^1}_{s{p^3}d\,\,{\text{hybridisation}}}$$       ( in first excitation state )

Hence, five $$s{p^3}d$$  hybrid orbitals are obtained. One orbital is already paired and rest four are overlapped with four $$p - {\text{orbitals}}$$   of four fluorine atoms on their axis in trigonal bipyramidal form.
This structure is distorted from trigonal bipyramidal to tetrahedral due to involvement of repulsion between lone pair and bond pair.

A molecule exists only if the bond order is positive. If bond order is zero or negative, the molecule does not exist.

No explanation is given for this question. Let's discuss the answer together.

(A) $$S{F_6} - $$  Octahedral
(B) $$SiC{l_4} - $$  Tetrahedral
(C) $$As{F_5} - $$  Trigonal bipyramidal
(D) $$BC{l_3} - $$  Trigonal planar

$$P$$ is related to $$s{p^3}$$ - hybridised $$C$$ - atom, $$Q$$ is related to $$s{p^2}$$ - hybridised $$C$$ - atom and $$R$$ is related to $$sp$$  - hybridised $$C$$ - atom.

In $$s{p^3}d,$$  the bond angle is $${120^ \circ }$$ and $${90^ \circ }.$$

Hybridisation of $${H_2}S{O_3} = \frac{1}{2}\left( {6 + 2 + 0} \right) = 4 = s{p^3}$$