Question

question 3 (1 point) retake question consider the following ions and elements: sn+2, sb+3, te-1, rb+1, i-1, ag what pair of ions are isoelectronic with each other? rb+1 and ag te-1 and i-1 sb+3 and i-1 sn+2 and sb+3 question 8 (1 point) retake question how many dots (nonbonding electrons) appear in the final lewis structure of phosphine (ph3)? 2 0 4 6

Explanation:

Step1: Determine electron - count for isoelectronic ions

Isoelectronic species have the same number of electrons. Sn (tin) has atomic number 50, so $Sn^{+2}$ has $50 - 2=48$ electrons. Sb (antimony) has atomic number 51, so $Sb^{+3}$ has $51 - 3 = 48$ electrons. Rb (rubidium) has atomic number 37, so $Rb^{+1}$ has $37-1 = 36$ electrons. Te (tellurium) has atomic number 52, so $Te^{-1}$ has $52 + 1=53$ electrons. I (iodine) has atomic number 53, so $I^{-1}$ has $53+1 = 54$ electrons. Ag (silver) has atomic number 47. So the pair $Sn^{+2}$ and $Sb^{+3}$ are isoelectronic.

Step2: Determine non - bonding electrons in $PH_3$

Phosphorus (P) has 5 valence electrons and hydrogen (H) has 1 valence electron. In $PH_3$, P is the central atom. The total number of valence electrons is $5+(3\times1)=8$. P forms 3 single bonds with 3 H atoms, using 6 electrons. So the number of non - bonding electrons is $8 - 6=2$.

Answer:

Question 3: Sn$^{+2}$ and Sb$^{+3}$
Question 8: 2