Which of the following statements is correct?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Cyanogen has the linear structure and it has short $C - C$ distance when compared to $C - C$ bond length in diamond

$C O > C O_{2} > C O_{3}^{2 -} :$ ‘Carbon – Oxygen’ bond distance in pm

The $N S i_{3}$ skeleton is pyramidal in $N {(S i H_{3})}_{3}$ whereas $P S i_{3}$ skeleton in $P {(S i H_{3})}_{3}$ is planar

Among tetrafluorides of $S, S e$ and $T e; S F_{4}$ is a solid, $S e F_{4}$ a gas and $T e F_{4}$ a liquid

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Cyanogen has the linear structure and the short C – C distance indicates onsiderable electron delocalisation.

$\begin{array}{l} 116 p m \\ N \equiv C - C \equiv N \\ 137 p m \end{array}$

(ii) In diamond C – C distance is 154 pm

(iii)

	$C - O$ bond order	$C - O$ bond length
$C O$	3	112.8 pm
$C O_{2}$	2	116 pm
$C O_{3}^{2 -}$	1.33	129 pm

(iv) The $N S i_{3}$ skeleton is planar in $N {(S i H_{3})}_{3}$ due to $p π - d π$ back bonding whereas $P S i_{3}$ skeleton in $P {(S i H_{3})}_{3}$ is pyramidal. $N - S i$ bond distance in $N {(S i H_{3})}_{3}$ is 173 pm and P – Si bond distance in $P {(S i H_{3})}_{3}$ is 225pm.