Welcome back Lawrence — missed you this week. The question describes a reaction where an element P with atomic number 11 reacts with an element Q with atomic number 17 to form a product PQ. We need to determine the number of electrons in P and Q in the product. Step 1: Determine the electron configuration and typical ion formation for P. Element P has an atomic number of 11, meaning it has 11 protons and 11 electrons in its neutral state. Its electron configuration is $1s^2 2s^2 2p^6 3s^1$. To achieve a stable octet, P will tend to lose its 1 valence electron, forming a positive ion with a charge of $+1$. $$ \text{P} \to \text{P}^+ + \text{e}^- $$ In the $\text{P}^+$ ion, the number of electrons will be $11 - 1 = 10$. Step 2: Determine the electron configuration and typical ion formation for Q. Element Q has an atomic number of 17, meaning it has 17 protons and 17 electrons in its neutral state. Its electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^5$. To achieve a stable octet, Q will tend to gain 1 electron, forming a negative ion with a charge of $-1$. $$ \text{Q} + \text{e}^- \to \text{Q}^- $$ In the $\text{Q}^-$ ion, the number of electrons will be $17 + 1 = 18$. Step 3: Determine the number of electrons in P and Q in the product PQ. When P and Q react to form the product PQ, P will form $\text{P}^+$ and Q will form $\text{Q}^-$. The compound formed is ionic, $\text{P}^+\text{Q}^-$. In the product PQ: • The number of electrons in P (as $\text{P}^+$) is 10. • The number of electrons in Q (as $\text{Q}^-$) is 18. Step 4: Compare with the given options. The number of electrons in P and Q in the product are 10 and 18 respectively. (A) 11 and 17 (B) 18 and 10 (C) 10 and 18 (D) 10 and 17 The correct option is (C). The final answer is $\boxed{\text{C}}$. Send me the next one 📸