Textbook Fundamentals J.7

[BKoh_2E]

Select an acid and a base for a neutralization reaction that results in the formation of (a) potassium bromide; (b) zinc nitrite;
(c) calcium cyanide, Ca(CN) 2 ; (d) potassium phosphate. Write the balanced equation for each reaction. I understand how to do part A, but I'm stuck on part B. I don't understand how they found the oxidation number for Zinc and the coefficients for the equation. Can somebody please give a step-by-explanation?

[Yuelai Feng 3E]

Hi! Zinc ion is usually 2+ charged because its electron configuration is [Ar]3d¹⁰4s², so it readily loses 2 valence electrons in 4s-orbital, similar to Group 2 metals.
Then, since hydroxide ion and nitrite ion are both 1- charged, the "zinc:anion" ratio is 1:2. This gives you Zn(OH)₂ and Zn(NO₂)₂.
Therefore the unbalanced equation is: Zn(OH)2 (aq) + HNO2 (aq) --> Zn(NO2)2 (aq) + H2O (l)
To balance the equation, you need to double the number of HNO2 on the left and H2O on the right: Zn(OH)2 (aq) + 2 HNO2 (aq) --> Zn(NO2)2 (aq) + 2 H2O (l)
Hope this helps!

[LeahSWM 2E]

Hi! Zinc ion is usually 2+ charged because its electron configuration is [Ar]3d¹⁰4s², so it readily loses 2 valence electrons in 4s-orbital, similar to Group 2 metals.
Then, since hydroxide ion and nitrite ion are both 1- charged, the "zinc:anion" ratio is 1:2. This gives you Zn(OH)₂ and Zn(NO₂)₂.
Therefore the unbalanced equation is: Zn(OH)2 (aq) + HNO2 (aq) --> Zn(NO2)2 (aq) + H2O (l)
To balance the equation, you need to double the number of HNO2 on the left and H2O on the right: Zn(OH)2 (aq) + 2 HNO2 (aq) --> Zn(NO2)2 (aq) + 2 H2O (l)
Hope this helps!

This helped me so much! thank you!