Q130 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Chemistry
Find the number of total atoms of all halogens present in 1 mole molecule of inter halogen compounds having $+7$ oxidation state of central halogen.
- Correct answerA$4.82\times10^{24}$
- B$4.21\times10^{24}$
- C$3.61\times10^{23}$
- D$3.01\times10^{23}$
Solution
The general formula of an interhalogen compound is $XX'_{n}$.
The oxidation state of the central halogen $X$ is $+n$.
Given that the oxidation state of the central halogen is $+7$, the formula of the compound is $XX'_{7}$ (for example, $IF_{7}$).
The number of halogen atoms in one molecule of $XX'_{7}$ is $1 + 7 = 8$.
The total number of halogen atoms in $1$ mole of the compound is $8 \times N_{A}$.
$8 \times 6.022 \times 10^{23} = 48.176 \times 10^{23} \approx 4.82 \times 10^{24}$
Answer: $4.82\times10^{24}$
The oxidation state of the central halogen $X$ is $+n$.
Given that the oxidation state of the central halogen is $+7$, the formula of the compound is $XX'_{7}$ (for example, $IF_{7}$).
The number of halogen atoms in one molecule of $XX'_{7}$ is $1 + 7 = 8$.
The total number of halogen atoms in $1$ mole of the compound is $8 \times N_{A}$.
$8 \times 6.022 \times 10^{23} = 48.176 \times 10^{23} \approx 4.82 \times 10^{24}$
Answer: $4.82\times10^{24}$
Hence, the correct answer is option: 'A'.