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← JEE Main 2022 · 26 Jul · Shift 2
ChemistryInteger+4 / -1Question 82 of 90

JEE Main 2022 · 26 Jul · Shift 2, Chemistry question 82

Consider an imaginary ion 2248X3−{}^{48}_{22}\mathrm X^{3-}. The nucleus contains 'a'% more neutrons than the number of electrons in the ion. The value of 'a' is ______. [nearest integer]

Answer and worked solution

Official answer: 4

Count. Neutrons 48−22=2648-22=26; electrons 22+3=2522+3=25.

Excess 26−2525=4%\frac{26-25}{25}=4\%.

Answer: 4