Question

answer the questions in the table below about the shape of the phosphodichloride (pci₂) anion. how many electron groups are around the central phosphorus atom? note: one electron group means one lone pair, one single bond, one double bond, or one triple bond. what phrase best describes the arrangement of these electron groups around the central phosphorus atom? (you may need to use the scrollbar to see all the choices.)

Explanation:

First Question (Number of Electron Groups):

Step 1: Determine valence electrons of P and Cl

Phosphorus (P) has 5 valence electrons, each Cl has 7, and the anion has a -1 charge (so +1 electron). So total valence electrons: \(5 + 2\times7 + 1 = 5 + 14 + 1 = 20\).

Step 2: Draw Lewis structure (simplified)

Central P, bonded to 2 Cl (single bonds, 2 electron groups). P has lone pairs: \( \frac{20 - 2\times2}{2} = 8\) (4 lone pairs? Wait, no: single bond is 2 electrons, so 2 bonds use 4 electrons. Remaining: \(20 - 4 = 16\), so lone pairs on P: \(16/2 = 8\)? Wait, no, P can have expanded octet? Wait, no, let's recalculate. Wait, the formula is \( \text{Total valence e}^- = \text{Valence e}^-(\text{P}) + 2\times\text{Valence e}^-(\text{Cl}) + \text{Charge}\). P: 5, Cl:7 each, charge -1 (so +1 e⁻). So \(5 + 2\times7 + 1 = 5 + 14 + 1 = 20\). Now, bonding: 2 single bonds (P - Cl), each uses 2 e⁻, so 4 e⁻. Remaining: \(20 - 4 = 16\) e⁻. These are lone pairs. On P: how many? Let's see, electron groups: each bond (2 bonds) and lone pairs. Wait, the note says electron group is lone pair, single, double, triple bond. So for \( \text{PCl}_2^- \), let's find the steric number (electron groups). Steric number = number of bonding groups + number of lone pairs on central atom. Let's calculate formal charge or use VSEPR. P is in group 5, so valence 5. In \( \text{PCl}_2^- \), P has 2 bonds (to Cl), so bonding electrons: 4, non - bonding: let's see, total valence e⁻:20. Bonding: 4, so non - bonding:16. But non - bonding on P: let's see, Cl atoms: each Cl has 3 lone pairs (since they are single bonded, 7 - 1 = 6, so 3 lone pairs each, 2 Cl: 6 lone pairs, 12 e⁻). So non - bonding on P: \(16 - 12 = 4\) e⁻, which is 2 lone pairs. Wait, no, that can't be. Wait, maybe I made a mistake. Wait, the correct way: for VSEPR, the central atom (P) has: number of bonding atoms (2 Cl) and lone pairs. Let's calculate the number of electron groups. The formula for electron groups: count all bonding (single, double, triple) and lone pairs. Let's use the formula for steric number (SN): \( \text{SN} = \frac{\text{Valence e}^-(\text{central}) - \text{Charge} + 2\times\text{Number of monovalent atoms}}{2} + \text{Number of double/triple bonds}\). Wait, no, better formula: \( \text{SN} = \text{Number of bonding groups} + \text{Number of lone pairs}\). For \( \text{PCl}_2^- \), P is central, Cl is monovalent (each contributes 1 bonding site). So \( \text{SN} = 2 + \text{lone pairs}\). The total valence e⁻:20. The bonding e⁻: 2×2 = 4 (since 2 single bonds). The non - bonding e⁻:20 - 4 = 16. These non - bonding e⁻ are distributed as lone pairs on Cl and P. Each Cl needs 6 non - bonding e⁻ (to complete octet), so 2 Cl: 12 e⁻. So non - bonding e⁻ on P:16 - 12 = 4, which is 2 lone pairs. So \( \text{SN}=2 + 2 = 4\)? Wait, no, that would be 4 electron groups. Wait, no, maybe I messed up the charge. Wait, the anion is \( \text{PCl}_2^- \), so charge is -1, meaning it has one extra electron. Let's re - calculate valence e⁻: P (5) + 2×Cl (7) + 1 (for -1 charge) = 5 + 14+1 = 20. Correct. Now, bonding: 2 single bonds (P - Cl), so 4 e⁻. Non - bonding:16 e⁻. Cl atoms: each Cl has 7 valence e⁻, in a single bond, they use 1 e⁻ for bonding, so 6 non - bonding (3 lone pairs) per Cl. So 2 Cl: 6×2 = 12 e⁻. So non - bonding on P:16 - 12 = 4 e⁻, which is 2 lone pairs. So electron groups on P: 2 (bonds) + 2 (lone pairs) = 4? Wait, no, that can't be, because 2 bonds and 2 lone pairs would be 4 electron groups. Wait, but maybe I made a mistake in the number of lone pairs on P. Wait, P can have an expanded octet, but let's check the VSEPR.…

Step 1: Recall VSEPR shapes for steric number 4

When the steric number (number of electron groups) is 4, the electron - group geometry (arrangement of electron groups) is tetrahedral. Because in VSEPR theory, for a steric number of 4, the electron groups arrange themselves in a tetrahedral geometry to minimize repulsion. The options given: linear (SN = 2), bent (SN = 3 or 4 with lone pairs), T - shaped (SN = 5), trigonal planar (SN = 3), trigonal pyramidal (SN = 4 with 1 lone pair), square planar (SN = 6 with 2 lone pairs), square pyramidal (SN = 6 with 1 lone pair), tetrahedral (SN = 4). Since our steric number is 4 (4 electron groups), the electron - group arrangement is tetrahedral.

First Question Answer:

4

Second Question Answer:

tetrahedral

So for the first blank (number of electron groups): 4. For the second blank (arrangement of electron groups): tetrahedral.

Answer:

Step 1: Recall VSEPR shapes for steric number 4

When the steric number (number of electron groups) is 4, the electron - group geometry (arrangement of electron groups) is tetrahedral. Because in VSEPR theory, for a steric number of 4, the electron groups arrange themselves in a tetrahedral geometry to minimize repulsion. The options given: linear (SN = 2), bent (SN = 3 or 4 with lone pairs), T - shaped (SN = 5), trigonal planar (SN = 3), trigonal pyramidal (SN = 4 with 1 lone pair), square planar (SN = 6 with 2 lone pairs), square pyramidal (SN = 6 with 1 lone pair), tetrahedral (SN = 4). Since our steric number is 4 (4 electron groups), the electron - group arrangement is tetrahedral.

First Question Answer:

4

Second Question Answer:

tetrahedral

So for the first blank (number of electron groups): 4. For the second blank (arrangement of electron groups): tetrahedral.