Question

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${CH}_3^{+}$ shows ${sp}^2$-hybridization whereas ${CH}_3^{-}$ show ${sp}^3$-hybridization.

• A. 1

Answer 1

The correct option is A 1
Carbon has the electron configuration of $1s^{2}2s^{2}2p^2$.

When one electron is promoted from $2sto2p_z$, then we have the configuration $2s^1,2p_x^1,2p_y^1,2p_z^1$ with four singly occupied atomic orbitals.

Each orbital is used to form one bond to $H$ in $C{H}_4$.

When we remove just the $H$ nucleus and leave behind the two electrons in the bond this gives $C{H}_3^{-}$, with $s{p}^3$ hybridization.

When we take away the $H^{-}$. This gives $C{H}_3^{+}$. Three bonds means three orbitals, one $s$ and two $p$. That means the third p orbital remains unhybridized. This situation is called $s{p}^2$ hybridization.

Hence $C{H}_3^{+}$ shows $s{p}^2$ hybridization whereas $C{H}_3^{-}$ show $s{p}^3$ hybridization.