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# Four elements P,Q, R and S have atomic numbers of 4, 10, 12 and 14 respectively ...

### Question

Four elements P,Q, R and S have atomic numbers of 4, 10, 12 and 14 respectively . Which of these electrons is a noble gas?

A)
P
B)
Q
C)
R
D)
S

### Explanation:

Noble gases are elements that have a full outer electron shell, making them very stable and unreactive. They belong to Group 18 (VIII) in the periodic table and include helium, neon, argon, krypton, xenon, and radon.

In the given question, we have four elements P, Q, R, and S with atomic numbers 4, 10, 12, and 14 respectively. To determine which of these elements is a noble gas, we need to identify the one with a full outer electron shell.

Let's analyze each element:

Element P: Atomic number 4 corresponds to beryllium (Be) which has an electron configuration of 1s²2s². Its outer shell has only 2 electrons, so it's not a noble gas.

Element Q: Atomic number 10 corresponds to neon (Ne) which has an electron configuration of 1s²2s²2p⁶. Its outer shell has 8 electrons, which is a full outer shell. Therefore, element Q is a noble gas.

Element R: Atomic number 12 corresponds to magnesium (Mg) which has an electron configuration of 1s²2s²2p⁶3s². Its outer shell has 2 electrons, so it's not a noble gas.

Element S: Atomic number 14 corresponds to silicon (Si) which has an electron configuration of 1s²2s²2p⁶3s²3p². Its outer shell has 4 electrons, so it's not a noble gas.

Based on our analysis, the correct answer is Option B: Q, as it has a full outer electron shell and is a noble gas.