Question

General electronic configuration of outermost and penultimate shell is $\left[Ne\right](n-1)s^2(n-1)p^6(n-1)d^{5}n{s}^2$. If $n=4$, then the number of protons in the nucleus will be:

• A. $>25$
• B. $>24$
• C. $25$
• D. $30$

Answer 1

The correct option is C $25$

Substituting the values, we get the configuration to be $\left[Ne\right]3s^{2}3p^{6}4s^{2}3d^5$.

The atomic number is $10+2+6+2+5=25$.

Number of proton = Atomic number

Therefore, the number of protons in the nucleus is $25.$

Hence, option C is correct.