Question

The successive ionization enthalpies of an element \(X\) in \(\text{kJ mol}^{-1}\) are \(1012,\ 1907,\ 2955,\ 4955,\ 6275\) and \(21260\) respectively, the element \(X\) is:

• A. C
• B. P
• C. S
• D. Cl

Hint:

A large jump in successive ionization enthalpy after removing \(n\) electrons indicates that the element has \(n\) valence electrons.

Answer 1

The Correct Option is B

Step 1: Observe the successive ionization enthalpies.
The given successive ionization enthalpies are \[ 1012,\ 1907,\ 2955,\ 4955,\ 6275,\ 21260\ \text{kJ mol}^{-1} \] We notice that the first five ionization enthalpies increase gradually.
But there is a very large jump between the fifth and sixth ionization enthalpies: \[ 6275 \rightarrow 21260 \]

Step 2: Interpret the large jump.
A sudden large jump in ionization enthalpy occurs when all valence electrons have been removed and the next electron has to be removed from an inner shell.
Since the large jump occurs after removing five electrons, the element must have \[ 5 \] valence electrons.

Step 3: Identify the group of the element.
An element with five valence electrons belongs to \[ \text{Group }15 \] Among the given options: \[ \text{C} \rightarrow \text{Group }14 \] \[ \text{P} \rightarrow \text{Group }15 \] \[ \text{S} \rightarrow \text{Group }16 \] \[ \text{Cl} \rightarrow \text{Group }17 \] Thus, the element is phosphorus: \[ X=\text{P} \]

Step 4: Final conclusion.
Therefore, the element \(X\) is \[ \boxed{\text{P}} \] Hence, the correct option is \[ \boxed{(2)} \]