Nonplanar cation is:

N \oplus H_{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\oplus {CH}_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{CH}_{3} \oplus C O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{({CH}_{3})}_{3} C^{+} {SbF}_{6}^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $N \oplus H_{4}$
Since, $N H_{4}^{⨁}$ (ammonium cation) has four bond pairs so has $s p^{3}$ hybridization resulting in tetrahedral geometry in which no plane of symmetry is there. So, no planar.

Suggest Corrections

Similar questions

(C H_{3})_{3} C - O - C H_{3}

(C H_{3}) C - I

C H_{3} - O H

(C H_{3})_{3} C - O H

C H_{3} - I

{(C H_{3})}_{3} C^{-}

C {H_{3}}^{-}

C H_{3} C {H_{2}}^{-}

{(C H_{3})}_{2} C H^{-}

$C H_{3} - B r M g - -- \to e t h e r (C H_{3})_{3} C - O H - -------- \to A (g a s) .^{'} A^{'} i s$

Functional isomer of $C H_{3} - C H_{2} - O H$ is:

Functional isomers of $C H_{3} - C H_{2} - C H_{2} - C O O H$

Join BYJU'S Learning Program