Question

Calculate the number of molecules in which central atom form bonds in excited state.

$ClF_3, ClF_5, CO_2, CCl_4, BF_3, SF_2, SF_4, NH_3, IF_7, PCl_5, PCl_3$

Answer 1

8

Answer 2

To determine the number of molecules in which the central atom forms bonds in an excited state, we need to examine the electron configuration of the central atom and the number of bonds it forms. A central atom forms bonds in an excited state if it needs to promote one or more electrons from its ground state configuration to higher energy orbitals to achieve the required number of unpaired electrons for bonding or to accommodate more than eight valence electrons (octet expansion).

Let's analyze each molecule:

1. $ClF_3$: Central atom is Cl. Ground state valence electron configuration of Cl is $3s^2 3p^5$. It forms 3 bonds and has 2 lone pairs. To form 3 bonds, Cl needs 3 unpaired electrons. In the ground state, Cl has only 1 unpaired electron ($3p^5$). To get 3 unpaired electrons, one electron from a paired $3p$ orbital is promoted to a vacant $3d$ orbital ($3s^2 3p^4 3d^1$), resulting in 3 unpaired electrons. Thus, Cl is in an excited state.
2. $ClF_5$: Central atom is Cl. Ground state valence electron configuration of Cl is $3s^2 3p^5$. It forms 5 bonds and has 1 lone pair. To form 5 bonds, Cl needs 5 unpaired electrons. In the ground state, Cl has 1 unpaired electron. To get 5 unpaired electrons, electrons are promoted from $3p$ and $3s$ orbitals to $3d$ orbitals ($3s^2 3p^3 3d^2$ or $3s^1 3p^3 3d^3$). Thus, Cl is in an excited state.
3. $CO_2$: Central atom is C. Ground state valence electron configuration of C is $2s^2 2p^2$. It forms 4 bonds (two double bonds). To form 4 bonds, C needs 4 unpaired electrons. In the ground state, C has 2 unpaired electrons ($2p^2$). To get 4 unpaired electrons, one electron from the $2s$ orbital is promoted to the vacant $2p$ orbital ($2s^1 2p^3$). Thus, C is in an excited state.
4. $CCl_4$: Central atom is C. Ground state valence electron configuration of C is $2s^2 2p^2$. It forms 4 single bonds. To form 4 bonds, C needs 4 unpaired electrons. In the ground state, C has 2 unpaired electrons. Promotion of a $2s$ electron to a $2p$ orbital ($2s^1 2p^3$) provides 4 unpaired electrons. Thus, C is in an excited state.
5. $BF_3$: Central atom is B. Ground state valence electron configuration of B is $2s^2 2p^1$. It forms 3 single bonds. To form 3 bonds, B needs 3 unpaired electrons. In the ground state, B has 1 unpaired electron ($2p^1$). Promotion of a $2s$ electron to a $2p$ orbital ($2s^1 2p^2$) provides 3 unpaired electrons. Thus, B is in an excited state.
6. $SF_2$: Central atom is S. Ground state valence electron configuration of S is $3s^2 3p^4$. It forms 2 bonds and has 2 lone pairs. To form 2 bonds, S needs 2 unpaired electrons. In the ground state, S has 2 unpaired electrons ($3p^4$). No electron promotion is required for bonding. Thus, S is in the ground state.
7. $SF_4$: Central atom is S. Ground state valence electron configuration of S is $3s^2 3p^4$. It forms 4 bonds and has 1 lone pair. To form 4 bonds, S needs 4 unpaired electrons. In the ground state, S has 2 unpaired electrons. Promotion of one electron from a paired $3p$ orbital to a vacant $3d$ orbital ($3s^2 3p^3 3d^1$) provides 4 unpaired electrons. Thus, S is in an excited state.
8. $NH_3$: Central atom is N. Ground state valence electron configuration of N is $2s^2 2p^3$. It forms 3 bonds and has 1 lone pair. To form 3 bonds, N needs 3 unpaired electrons. In the ground state, N has 3 unpaired electrons ($2p^3$). No electron promotion is required for bonding. Nitrogen is a second-period element and cannot expand its octet. Thus, N is in the ground state.
9. $IF_7$: Central atom is I. Ground state valence electron configuration of I is $5s^2 5p^5$. It forms 7 bonds. To form 7 bonds, I needs 7 unpaired electrons. In the ground state, I has 1 unpaired electron ($5p^5$). Promotion of electrons from $5p$ and $5s$ orbitals to vacant $5d$ orbitals ($5s^1 5p^3 5d^3$) provides 7 unpaired electrons. Thus, I is in an excited state.
10. $PCl_5$: Central atom is P. Ground state valence electron configuration of P is $3s^2 3p^3$. It forms 5 bonds. To form 5 bonds, P needs 5 unpaired electrons. In the ground state, P has 3 unpaired electrons ($3p^3$). Promotion of one electron from the $3s$ orbital to a vacant $3d$ orbital ($3s^1 3p^3 3d^1$) provides 5 unpaired electrons. Thus, P is in an excited state.
11. $PCl_3$: Central atom is P. Ground state valence electron configuration of P is $3s^2 3p^3$. It forms 3 bonds and has 1 lone pair. To form 3 bonds, P needs 3 unpaired electrons. In the ground state, P has 3 unpaired electrons ($3p^3$). No electron promotion is required for bonding. Thus, P is in the ground state.

The molecules in which the central atom forms bonds in an excited state are: $ClF_3, ClF_5, CO_2, CCl_4, BF_3, SF_4, IF_7, PCl_5$.

Counting these molecules, we find there are 8.