Question

Version 1 11) Which of the following species would be paramagnetic? A) Fest B) Sr C) 02 D) Mg E) Zn 12) Place the following in order of increasing radius. Sre Se? Br A) Br <Sr? <Se B) Sr- <Set- <Br C) Se?- < Br<Sr? D) Br <Set- <Sr E) Sr* <Br <Se? 13) Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2 HAO-(I) - 2 HAOI) +O:(g) AH = -196 kg Calculate the value of q (kJ) in this exothermic reaction when 4.60 g of hydrogen peroxide decomposes at constant pressure. A) -13.3 kJ B) -26.5 kJ C) 1.25 kJ D) -0.0189 kJ E) -2.65 x 10 kJ 14) Choose the best Lewis structure for CH:Cl. =c= A) B) H-H-2-0-

Answer 1

11. A species is said to be paramagnetic if it has unpaired electrons in its outermost shell.

a) Electronic configuration of Fe = 3d⁶ 4s²

Electronic configuration of Fe³⁺ = 3d⁵

Each d- orbital will have 1 unpaired electron and a total of 5 unpaired electrons. So, it will be paramagnetic.

b) Electronic configuration of Sr = 4p⁶ 5s²

Electronic configuration of Sr²⁺ = 4p⁶

There are no unpaired electrons, so it will be diamagnetic.

c) electronic configuration of O = 2p⁴

Electronic configuration of O^2- = 2p⁶

There are also no unpaired electrons, so it will be diamagnetic.

d) electronic configuration of Mg = 3s²

The electrons are paired and hence, it is diamagnetic

e) electronic configuration of Zn = 3d¹⁰ 4s²

Electronic configuration of Zn²⁺ = 3d¹⁰

So, this is also diamagnetic.

Hence, Fe³⁺ is paramagnetic among given species.

12. For comparing ionic radius, first check whether the species are isoelectronic (containing same number of electrons).

In this case, all the given species are isoelectronic containing 36 electrons.

In such a case ionic radius varies as: Anion > Neutral species > Cation

And, ionic radius of more charged anionic species > ionic radius of less charged anionic species.

Ionic radius of less charged cationic species > ionic radius of more charged cationic species.

Therefore, order of ionic radius: Se^2- > Br^- > Sr²⁺.

Option E is correct.

13.

For decomposing, a mol of Hy be heat required = -196 kJ * Fox & mol x 349/mol of Hon heat required = -196 kJ 3 689 of Hy hey heat required = -196 kJ 3 lg g Halay heat required = -196 kJ = For 4.60g of Hd, heat required - 196 x 4.60 kJ 68 68 = -13.3 k 1 option A is correct.