Arrange in decreasing order, the energy of 2s orbital in the following atoms H, Li, Na, K

Question

Arrange in decreasing order, the energy of 2s orbital in the following atoms H, Li, Na, K (A) E2 s( H )>E2 s( Li )>E2 s( Na )>E2 s( K ) (B) E2 s( H )>E2 s( Na )>E2 s( Li )>E2 s( K ) (C) E2 s( H )>E2 s( Na )=E2 s( K )>E2 s( Li ) (D) E2 s( K )< E2 s( Na )< E2 s( Li )< E2 s( H )

Tardigrade Admin · Accepted Answer

The electronic configuration of the given atoms is as

H = 1 s^{1}

L i = 1 s^{2}, 2 s^{1}

N a = 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{1}

K = 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}, 4 s^{1}

Thus, the increasing order of energy of

2 s

orbitals is

E_{2 s (K)} < E_{2 s (N a)} < E_{2 s (L i)} < E_{2 s (H)}

Q. Arrange in decreasing order, the energy of $2 s$ orbital in the following atoms $H, L i, N a, K$

$E_{2 s (H)} > E_{2 s (L i)} > E_{2 s (N a)} > E_{2 s (K)}$

$E_{2 s (H)} > E_{2 s (N a)} > E_{2 s (L i)} > E_{2 s (K)}$

$E_{2 s (H)} > E_{2 s (N a)} = E_{2 s (K)} > E_{2 s (L i)}$

$E_{2 s (K)} < E_{2 s (N a)} < E_{2 s (L i)} < E_{2 s (H)}$

Q. Arrange in decreasing order, the energy of 2s orbital in the following atoms H,Li,Na,K

E2s(H)​>E2s(Li)​>E2s(Na)​>E2s(K)​

E2s(H)​>E2s(Na)​>E2s(Li)​>E2s(K)​

E2s(H)​>E2s(Na)​=E2s(K)​>E2s(Li)​

E2s(K)​<E2s(Na)​<E2s(Li)​<E2s(H)​

The electronic configuration of the given atoms is as H=1s1 Li=1s2,2s1 Na=1s2,2s22p6,3s1 K=1s2,2s22p6,3s23p6,4s1 Thus, the increasing order of energy of 2s orbitals is E2s(K)​<E2s(Na)​<E2s(Li)​<E2s(H)​

Q. Arrange in decreasing order, the energy of $2 s$ orbital in the following atoms $H, L i, N a, K$

$E_{2 s (H)} > E_{2 s (L i)} > E_{2 s (N a)} > E_{2 s (K)}$

$E_{2 s (H)} > E_{2 s (N a)} > E_{2 s (L i)} > E_{2 s (K)}$

$E_{2 s (H)} > E_{2 s (N a)} = E_{2 s (K)} > E_{2 s (L i)}$

$E_{2 s (K)} < E_{2 s (N a)} < E_{2 s (L i)} < E_{2 s (H)}$