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?

Answer Details

A noble gas is an element with a complete outermost electron shell. The outermost shell of an atom can contain up to 8 electrons. The atomic number of P is 4, which means it has 4 electrons. The electron configuration of P is 1s2 2s2, which indicates that it has two electrons in the first shell and two electrons in the second shell. Since the outermost shell is not complete, P is not a noble gas. The atomic number of Q is 10, which means it has 10 electrons. The electron configuration of Q is 1s2 2s2 2p6, which indicates that it has two electrons in the first shell, two electrons in the second shell, and six electrons in the third shell. Since the outermost shell of Q is complete with 8 electrons, Q is a noble gas (specifically, it is neon). The atomic number of R is 12, which means it has 12 electrons. The electron configuration of R is 1s2 2s2 2p6 3s2 3p6, which indicates that it has two electrons in the first shell, two electrons in the second shell, six electrons in the third shell, and two electrons in the fourth shell. Since the outermost shell of R is not complete, R is not a noble gas. The atomic number of S is 14, which means it has 14 electrons. The electron configuration of S is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2, which indicates that it has two electrons in the first shell, two electrons in the second shell, six electrons in the third shell, ten electrons in the fourth shell, and two electrons in the fifth shell. Since the outermost shell of S is not complete, S is not a noble gas. Therefore, the answer is (B) Q, which is neon and has a complete outermost electron shell.