Arrange the carbanions in decreasing order of stability (usual NCERT-level argument based on hybridisation and alkyl substitution): $HC \equiv C X^{-}$ , $CH X_{2} = CH X^{-}$ , $CH X_{3} CH X_{2} X^{-}$ , $(CH X_{3}) X_{3} C X^{-}$

A $HC \equiv C X^{-} > CH X_{3} CH X_{2} X^{-} > CH X_{2} = CH X^{-} > (CH X_{3}) X_{3} C X^{-}$
B $HC \equiv C X^{-} > CH X_{2} = CH X^{-} > CH X_{3} CH X_{2} X^{-} > (CH X_{3}) X_{3} C X^{-}$
C $CH X_{2} = CH X^{-} > HC \equiv C X^{-} > CH X_{3} CH X_{2} X^{-} > (CH X_{3}) X_{3} C X^{-}$
D $(CH X_{3}) X_{3} C X^{-} > CH X_{3} CH X_{2} X^{-} > CH X_{2} = CH X^{-} > HC \equiv C X^{-}$

Answer: B

Stability increases with s-character of the orbital holding the lone pair:

s p

(50%) >

s p^{2}

(33%) >

s p^{3}

(25%). Among the two

s p^{3}

anions, the tertiary one is destabilised by three +I alkyl groups, so ethyl is more stable than tert-butyl. This gives the stated order.