Question 42 of 62intermediate🔧 ApplyNumerical2 marks

An ion with mass number 37 possesses one unit of negative charge. If the ion conatins 11.1% more neutrons than the electrons, find the symbol of the ion.

Correct Answer

The symbol of the ion is 1737Cl−^{37}_{17}\text{Cl}^{-}.

Exercise: EXERCISES | Q: 2.43 | (Page 41)
For More Understanding

Explanation

The ion has a mass number of 37 and a single negative charge, meaning it has one more electron than protons. By setting up an equation where the number of neutrons (mass number minus atomic number) is 11.1% greater than the number of electrons (atomic number plus one), we can solve for the atomic number. An atomic number of 17 corresponds to chlorine, yielding the final symbol.

Solution Steps

  1. Step 1: Let the atomic number (number of protons) be ZZ. Since the ion has one unit of negative charge, the number of electrons is Z+1Z + 1. The mass number A=37A = 37, so the number of neutrons n=37−Zn = 37 - Z.

  2. Step 2: The ion contains 11.1% more neutrons than electrons. Since 11.1% is approximately 19\frac{1}{9}, we can write: n=electrons+19×electrons=109×(Z+1)n = \text{electrons} + \frac{1}{9} \times \text{electrons} = \frac{10}{9} \times (Z + 1).

  3. Step 3: Equating the two expressions for neutrons: 37−Z=109(Z+1)37 - Z = \frac{10}{9} (Z + 1). Solving this gives 333−9Z=10Z+10333 - 9Z = 10Z + 10, which simplifies to 19Z=32319Z = 323, so Z=17Z = 17.

  4. Step 4: The atomic number Z=17Z = 17 corresponds to the element Chlorine (Cl). The number of neutrons is 37−17=2037 - 17 = 20, and the number of electrons is 17+1=1817 + 1 = 18. Therefore, the symbol of the ion is 1737Cl−^{37}_{17}\text{Cl}^{-}.

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