Question

question 17 (0.5 points) give the formula from the systematic name molybdenum (iii) hydrogen phosphite 1 2 a. mo2 b. mo c. mo3 d. (cn)2 e. so3 f. (so3)2 g. (c2h3o2)2 h. c2h3o2 i. hso3 j. c2 k. c3 l. po3 m. (hpo3)2 n. h2po3 o. oh p. (oh)2 q. s r. so4 s. (so4)2 t. hpo3 u. (hpo3)3 p. moliy

Explanation:

Step1: Identify ion charges

Molybdenum(III) has a charge of +3, written as $Mo^{3 + }$. Hydrogen phosphite ion has a charge of - 2, written as $HPO_{3}^{2 - }$.

Step2: Determine formula by charge - balance

To balance the charges, we need 2 $Mo^{3+}$ ions and 3 $HPO_{3}^{2 - }$ ions. So the formula is $Mo_{2}(HPO_{3})_{3}$. But looking at the options, we consider the correct components for each part. The symbol for molybdenum is Mo and for hydrogen phosphite - related we consider the $HPO_{3}$ part. Since we need to represent the combination based on the ions, for molybdenum we just have Mo as the element symbol and for hydrogen phosphite in the correct form related to the ion combination we consider $(HPO_{3})_{3}$. But among the options, we note that we first pick the Mo symbol for molybdenum and then the correct hydrogen - phosphite related option. The symbol for molybdenum is Mo (option B) and the correct hydrogen phosphite - related option considering the ion is $HPO_{3}$ (option T).

Answer:

B. Mo, T. $HPO_{3}$