Question

The increasing order of \(2s\)-orbital's of the following atoms with respect to energy is \[ H,\; K,\; Li,\; Na \]

• A. \(E_{2s(H)}\lt E_{2s(Li)}\lt E_{2s(Na)}\lt E_{2s(K)}\)
• B. \(E_{2s(Li)}\lt E_{2s(H)}\lt E_{2s(Na)}\lt E_{2s(K)}\)
• C. \(E_{2s(H)}\lt E_{2s(Li)}\lt E_{2s(K)}\lt E_{2s(Na)}\)
• D. \(E_{2s(K)}\lt E_{2s(Na)}\lt E_{2s(Li)}\lt E_{2s(H)}\)

Hint:

The energy of inner-shell orbitals increases with increasing atomic number due to increasing effective nuclear charge. For the given atoms: \[ 2s(H)\lt 2s(Li)\lt 2s(Na)\lt 2s(K) \]

Answer 1

The Correct Option is A

Step 1: Decide what controls 2s energy.
For a many-electron atom, the orbital energy is set by how strongly the nucleus pulls that electron, that is the effective nuclear charge. A stronger pull makes the energy more negative, hence lower.

Step 2: Treat hydrogen.
Hydrogen has just one proton, so its $2s$ electron feels the smallest pull and sits at the lowest (most negative) energy. So $H$ comes first.

Step 3: Compare the alkali metals.
For Li, Na and K the $2s$ orbital is an inner shell. As we go down Li to Na to K, the nuclear charge rises, so the inner $2s$ electron is pulled in harder and its energy becomes more negative.

Step 4: Order the alkali metals.
Bigger nuclear charge gives lower $2s$ energy, so $E_{2s}(K) < E_{2s}(Na) < E_{2s}(Li)$.

Step 5: Place hydrogen.
Since hydrogen's pull is weakest, its $2s$ is the highest of all, sitting above Li.

Step 6: Write increasing order.
Increasing energy (least negative on the right): $E_{2s}(K) < E_{2s}(Na) < E_{2s}(Li) < E_{2s}(H)$.
\[ \boxed{E_{2s(K)} < E_{2s(Na)} < E_{2s(Li)} < E_{2s(H)}} \]