Consider the species CH_4,NH_4^+ and BH_4^-. | vedclass

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(B) $1$. Calculate the total number of electrons in each species:$CH_4$: $6 + 4(1) = 10$ electrons.$NH_4^+$: $7 + 4(1) - 1 = 10$ electrons.$BH_4^-$: $

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They are isoelectronic and all have tetrahedral structures

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(B) $1$. Calculate the total number of electrons in each species:$CH_4$: $6 + 4(1) = 10$ electrons.$NH_4^+$: $7 + 4(1) - 1 = 10$ electrons.$BH_4^-$: $5 + 4(1) + 1 = 10$ electrons.Since all species have $10$ electrons,they are isoelectronic.$2$. Determine the geometry:All three species have a central atom bonded to $4$ hydrogen atoms with no lone pairs on the central atom. According to $VSEPR$ theory,they all exhibit $sp^3$ hybridization and possess a tetrahedral geometry.

Consider the species $CH_4$,$NH_4^+$ and $BH_4^-$. Choose the correct option with respect to these species.

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