Question

directions: drag and drop the correct answer choice to each answer blank. arrange the elements whose electron configurations are given below in increasing order of atomic size. element c 1s22s22p63s1 element a 1s22s22p6 element b 1s22s2 element d 1s22s22p3

Explanation:

Step1: Determine the atomic number from electron - configurations

The number of electrons in a neutral atom is equal to its atomic number. For Element A: \(1s^{2}2s^{2}2p^{6}\), the number of electrons \(=2 + 2+6=10\), atomic number \(Z = 10\) (Neon). For Element B: \(1s^{2}2s^{2}\), the number of electrons \(=2 + 2 = 4\) (Beryllium). For Element C: \(1s^{2}2s^{2}2p^{6}3s^{1}\), the number of electrons \(=2+2 + 6+1=11\) (Sodium). For Element D: \(1s^{2}2s^{2}2p^{3}\), the number of electrons \(=2+2 + 3=7\) (Nitrogen).

Step2: Apply the periodic - trend of atomic size

Atomic size increases down a group and decreases across a period. In the periodic table, Beryllium (B) is in Group 2 and Period 2, Nitrogen (D) is in Group 15 and Period 2, Neon (A) is in Group 18 and Period 2, and Sodium (C) is in Group 1 and Period 3. Since Sodium is in the third - period and the other elements are in the second - period, Sodium has the largest atomic size. Among Beryllium, Nitrogen, and Neon in the second period, atomic size decreases from left to right. So, Beryllium has a larger atomic size than Nitrogen, and Nitrogen has a larger atomic size than Neon.

Answer:

Element A < Element D < Element B < Element C