Question

Rank the following compounds from largest to smallest according to their expected magnitude in lattice energy. Rank from largest to smallest lattice energy To rank items as equivalent, overlap them. a. Cs1 b. KI c. MgS d. KBr e. BeO Largest Lattice Energy Smallest Lattice Energy

Answer 1

To rank the compounds (BeO, MgS, KI, KBr, and CsI) from largest to smallest lattice energy, the order is BeO > MgS > KI > KBr > CsI, with BeO having the highest due to small, highly charged ions, and CsI the lowest due to larger, singly charged ions.

Step-by-step explanation:

To rank the given compounds from largest to smallest according to their expected magnitude in lattice energy, we consider two main factors: the charge on the ions and the size of the ions. The higher the charge and the smaller the size of the ions, the larger the lattice energy. Based on these criteria:

1. BeO (Be²⁺ and O²⁻)
2. MgS (Mg²⁺ and S²⁻)
3. CaI (K⁺ and I⁻)
4. KBr (K⁺ and Br⁻)
5. CaI (Cs⁺ and I⁻)

BeO has the largest lattice energy because it contains the smallest ions with the highest charges. MgS follows, having ions with the same charge but larger sizes. KI, KBr, and CsI have lower lattice energies due to having ions with ± charges, and among them, the size of the ions increases from KI to KBr to CsI.

Answer 2

BeO, MgS, KBr, KI, CsI

Step-by-step explanation:

Hello,

The lattice energy is defined as "a measure of the energy contained in the crystal lattice of a compound, equal to the energy that would be released if the component ions were brought together from infinity", therefore it will be larger in compounds composed by ions with high charge and ions with small radius .

Thus, the largest is BeO (2+ 2- and both of them are small), next MgS (2+ 2- as the BeO, nonetheless it has bigger ions), afterwards KBr (1+ 1- considering Br- to be smaller than I-) , then KI [1+ 1 considering I- larger than Br- and K+ smaller than Cs+] and finally CsI.

Best regards.