Question

question 4 (0.25 points)listenwhat is the electron configuration of a cobalt (ii) ion?$ar4s^{2}4d^{7}LXB0ar4s^{2}3d^{5}LXB1ar4s^{2}3d^{10}4p^{5}LXB2ar4s^{2}4d^{10}4p^{6}$question 6 (0.25 points)listenin what block, group, and period on the periodic table can the atom with a $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$ ground-state electron configuration be found?p-block, group 7a, and period 3p-block, group 7a, and period 4p-block, group 7a, and period 5p-block, group 6a, and period 4

Explanation:

Question 4

Step1: Find neutral Co electron count

Cobalt (Co) has an atomic number of 27, so neutral Co has 27 electrons. Its electron configuration is $[Ar]4s^23d^7$.

Step2: Remove 2 electrons for Co(II)

For transition metals, 4s electrons are lost first. Remove 2 electrons from the 4s orbital: $[Ar]3d^7$.

Question 5

Step1: Find neutral Br electron count

Bromine (Br) has an atomic number of 35, so neutral Br has 35 electrons. Its electron configuration is $[Ar]4s^23d^{10}4p^5$.

Step2: Add 1 electron for Br⁻

A bromide ion ($Br^-$) gains 1 electron, filling the 4p orbital: $[Ar]4s^23d^{10}4p^6$.

Question 6

Step1: Identify highest energy level

The outermost principal energy level is $n=4$, so the period is 4.

Step2: Identify valence electrons

Valence electrons are $4s^24p^5$, total 7 valence electrons, so group 7A. The last electron fills a p orbital, so it is in the p-block.

Answer:

Question 4: [Ar]3d⁷
Question 5: [Ar]4s²3d¹⁰4p⁶
Question 6: p-block, group 7A, and period 4