On the basis of electronic configuration the period and group of `B_(5)^(9)` is

Options

A. 2 and IIIA
B. 3 and IIA
C. 4 and VIA
D. 5 and VIIA

Explanation

To determine the period and group of the element with the atomic number \( B_5^9 \) (which indicates that it has 5 protons and 5 electrons), we will follow these steps:

### Step 1: Identify the Atomic Number
The atomic number of the element is 5. This means it has 5 electrons.

### Step 2: Write the Electronic Configuration
The electronic configuration for an element is determined by filling the electron shells according to the Aufbau principle. The first shell (1s) can hold 2 electrons, the second shell (2s) can hold 2 electrons, and the second shell also has a 2p subshell that can hold up to 6 electrons.

- The first 2 electrons will fill the 1s subshell: \( 1s^2 \)
- The next 2 electrons will fill the 2s subshell: \( 2s^2 \)
- The remaining 1 electron will go into the 2p subshell: \( 2p^1 \)

Thus, the electronic configuration of the element is:
\[ 1s^2 \, 2s^2 \, 2p^1 \]

### Step 3: Determine the Period
The period of an element corresponds to the highest principal quantum number (n) in its electronic configuration. In our case, the outermost shell is the second shell (2), as indicated by the \( 2s \) and \( 2p \) subshells.

Therefore, the period number is:
- **Period = 2**

### Step 4: Determine the Group
To find the group of the element, we look at the number of electrons in the outermost shell. The outermost shell (n=2) contains:
- 2 electrons in the 2s subshell
- 1 electron in the 2p subshell

This totals to 3 electrons in the outermost shell. In the periodic table, elements with 3 electrons in their outermost shell belong to Group 13 (or IIIA).

Therefore, the group number is:
- **Group = 13 (or IIIA)**

### Final Answer
Based on the electronic configuration, the element \( B_5^9 \) is in:
- **Period: 2**
- **Group: 13 (or IIIA)**

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