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Chapter 8: Problem 171

A compound, \(\mathrm{XF}_{5},\) is 42.81\(\%\) fluorine by mass. Identify the element \(\mathrm{X}\) . What is the molecular structure of \(\mathrm{XF}_{5} ?\)

Short Answer

Expert verified
The unknown element X is iodine (I), and the molecular structure of IF5 (iodine pentafluoride) is square pyramidal.

Step by step solution

01

Determine the mass percentage of the unknown element X #

Given that the compound XF5 is 42.81% fluorine by mass, we can find the mass percentage of the unknown element X by subtracting the mass percentage of fluorine from 100%: Mass percentage of X = 100% - 42.81% = 57.19%
02

Calculate the molar mass of XF5 #

To find the molar mass of XF5, we will use the molar mass of fluorine, which is 18.998 g/mol. There are 5 atoms of fluorine in XF5, so the total mass of fluorine is: Mass of 5F = 5 × 18.998 g/mol = 94.99 g/mol Now, let's denote the molar mass of X as M, then we can use the mass percentage to find the molar mass of XF5: Mass of X = 0.5719 × (M + 94.99 g/mol)
03

Determine the atomic mass of X #

Since we know the mass of 5F (94.99 g/mol), we can use the mass percentage of X to calculate the mass of X, which is 0.5719 of the molar mass of XF5: Mass of X = 0.4281 × (M + 94.99 g/mol) Multiply both sides of the equation by (M + 94.99 g/mol): M + 94.99 g/mol = 0.5719 × (M + 94.99 g/mol) / 0.4281 Solve for M: M ≈ 137.33 g/mol The atomic mass closest to 137.33 g/mol is 137.33 g/mol, which is the atomic mass of iodine (I). Thus, the unknown element X is iodine (I).
04

Predict the molecular structure of XF5 #

Now that we know the element X is iodine (I), we can use the VSEPR theory to predict the molecular structure of IF5. The central iodine atom has 7 valence electrons, and each of the fluorine atoms contributes a single electron. In IF5, there are 5 bonding electron pairs and 1 lone pair of electrons. According to VSEPR theory, this arrangement results in a square pyramidal molecular structure. #Answer#: The unknown element X is iodine (I), and the molecular structure of IF5 (iodine pentafluoride) is square pyramidal.

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Most popular questions from this chapter

Predict the molecular structure, bond angles, and polarity (has a net dipole moment or has no net dipole moment) for each of the following compounds. a. \(\mathrm{SeCl}_{4} \quad\) d. \(\mathrm{CBr}_{4}\) b. \(\mathrm{SO}_{2} \quad\) e. \(\mathrm{IF}_{3}\) c. \(\mathrm{KrF}_{4} \quad\) f. \(\mathrm{ClF}_{5}\)

Draw a Lewis structure for the \(N, N\) -dimethylformamide molecule. The skeletal structure is Various types of evidence lead to the conclusion that there is some double bond character to the C-N bond. Draw one or more resonance structures that support this observation.

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part c, why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not $\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?$ What about MgO compared to \(\mathrm{MgO}_{2} ? \mathrm{Mg}_{2} \mathrm{O} ?\)

Use the following standard enthalpies of formation to estimate the \(\mathrm{N}-\mathrm{H}\) bond energy in ammonia: $\mathrm{N}(g), 472.7 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}(g),\( \)216.0 \mathrm{kJ} / \mathrm{mol} ; \mathrm{NH}_{3}(g),-46.1 \mathrm{kJ} / \mathrm{mol}$ . Compare your value to the one in Table \(8.5 .\)

Give an example of an ionic compound where both the anion and the cation are isoelectronic with each of the following noble gases. a. \(\mathrm{Ne} \quad\) c. \(\mathrm{Kr}\) b. \(\mathrm{Ar} \quad\) d. \(\mathrm{Xe}\)
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