Question

The electronegativity of the followingelements increases in the order

• A. C, N, Si, P
• B. N, Si, C, P
• C. Si, P, C, N
• D. P, Si, N, C

Answer 1

The correct option is C

Si, P, C, N

The elements in the question are:

C $\to$ 1$s^2$ 2$s^2$ 2$p^2$ ($2^{nd}$ period)

N $\to$ 1$s^2$2$s^2$2$p^3$ ($2^{nd}$ period)

Si $\to$ 1$s^2$2$s^2$2$p^6$3$s^2$3$p^2$ ($3^{rd}$ period)

P $\to$ 1$s^2$2$s^2$2$p^6$3$s^2$3$p^3$ ($3^{rd}$ period)

We know that,

In a period moving from left to right, the electronegativity increases due to the increase in effective nuclear charge.

Ina group moving from top to bottom, the electronegativitiy decreases because atomic radius increases.

because

Atomic Radius: As the atomic radius of the element increases the electronegativity value decreases.

Electronegativity $\alpha$ $\frac{1}{atomicradius}$

and

Effective Nuclear Charge: The electronegativity value increases as the effective nuclear charge on the atomic nucleus increases. Electronegativity ∝Effective nuclear charge ($Z_{eff}$)

Now, if we look at the atomic radius of these elements.

Si > P > C > N (Atomic radius)

Therefore,

N > C > P > Si (Electronegativity)

$\Rightarrow$ I hope, now you are clear with the idea.