Two metals A and B possess the same number of electrons in the L shell and they differ by 1 unit in the number of electrons present in valence shell and number of valence electrons is greater in B than A. The ions formed get the configuration of the nearest inert gases namely argon and neon, respectively.Give the probable electronic configurations of the metals corresponding to A and B and also give the formulae of their corresponding nitrite, nitrate, sulphite and bisulphite.

Explanation

To solve the problem, we need to determine the electronic configurations of metals A and B based on the information provided and then derive the formulas for their corresponding nitrite, nitrate, sulphite, and bisulphite.

### Step 1: Determine the Electronic Configurations of Metals A and B

1. **Understanding the Shells**:
- Both metals A and B have the same number of electrons in the L shell.
- The K shell can hold a maximum of 2 electrons, and the L shell can hold a maximum of 8 electrons.

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