Question

which of the following elements has the following shorthand electron configuration? ar 4s²3d⁷
cobalt
calcium
copper
carbon
which shorthand electron configuration is written correctly? kr 5s² 4d¹⁰5p³ xe 5s¹ 4d¹⁰5p⁶
option 4
option 3

Explanation:

Step1: Recall electron - configuration rules

The noble - gas core [Ar] represents 18 electrons. Adding the electrons from the outer - shell configuration $4s^{2}3d^{7}$ (2 + 7 = 9 more electrons), the total number of electrons is 18+9 = 27.

Step2: Identify the element by atomic number

The element with atomic number 27 is cobalt.

For the second part:

Step1: Analyze electron - configuration rules

In a correct electron - configuration, the orbitals should be filled in the order of increasing energy. The configuration [Kr] $5s^{2}4d^{10}5p^{3}$ is correct as it follows the Aufbau principle. The configuration [Xe] $5s^{1}4d^{10}5p^{6}$ is incorrect because if we start with the [Xe] core (54 electrons), adding 1 + 10+6 = 17 more electrons gives an incorrect electron count and also violates the normal filling order.

Answer:

First question: A. Cobalt
Second question: Option 4. [Kr] $5s^{2}4d^{10}5p^{3}$