The ratio of atomic numbers of two elements X and Y is 1:7. The radius of the third orbit of single electron species of X is $1.587 Å .$ Identify the penultimate shell of Y.

N-shell

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M-shell

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L-shell

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O-shell

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Solution

The correct option is B
M-shell

Explanation for correct answer
The radius of n^th orbit is calculated by the following formula,
$r_{n} = \frac{0.529 \times n^{2}}{Z}$
Here, n is the orbit number and Z is the atomic number.
Substitute the values,
$\begin{matrix} 1.587 Å & = & \frac{0.529 \times 3^{2}}{Z} \\ Z & = & 3 \end{matrix}$
Therefore, the ratio of atomic numbers of two elements X and Y is 3 : 21.
The shell immediately preceding the valence or the ultimate shell is known as the penultimate shell. Specifically, the atom's second last shell.
Each shell can only hold a certain number of electrons: the first shell (K) can only hold two electrons, the second shell (L) can only hold eight (2 + 6) electrons, the third shell (M) can only hold 18 (2 + 6 + 10) electrons, and so on.
So, the penultimate shell of Y is M (2, 8, 11).
Hence, the correct option is M.

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The ratio of atomic numbers of two elements X and Y is 1:7. The radius of the third orbit of single electron species of X is 1.587Å. Identify the penultimate shell of Y.

The ratio of atomic numbers of two elements X and Y is 1:7. The radius of the third orbit of single electron species of X is $1.587 Å .$ Identify the penultimate shell of Y.