Question

3.20 Given the following data for the length of a side of the unit cell for compounds that crystallize in the rock-salt structure, determine the cation radii: MgSe (545 pm), CaSe (591 pm), SrSe (623 pm), Base (662 pm). (Hint: To determine the radius of Se--, assume that the Se2-ions are in contact in MgSe.)

My question is how is the anion length of 193pm determined? This question has already been posted and solved on this site twice. I see the formula but need a breakdown on exactly how 193 pm is determined from the given length of 545pm. I don't understand the formula shown in the answer provided. Thank you.

Answer 1

Question : Mose having rock salt type structure and we all knew that en rock salt anion will occupy can corner and of every cube se will form FCC - ABC-ABC-~ABG. - tybe system faces = Se? ne Otto ( 8 carner +6 faces Pleit (8x1/9 + 6x13). from Picture I = mgat C position of octer hedral void ) = 12 edge one center . = 12x1/4+1 set = 4 mg 24 mgat 502- ser- ar a=unit cell length 1 = 545pm a= rsp?t Smght

rmg2t, &sez radius of cation and anion from Picture of octahedral void. O 4 AB= rmg2+ +05e2-cate AC=28se 2- cas 45 = AC = coonco AB - rngat trse 2- 8rse2- &mg2t trse2- = /28se 2- rmg2t = (1.414-1)=-414 Vse2- radicy ratio Trmget=.414 / rule rse2- rmg?t=41488e2- rmg2t +8se2 = 27215= 9 yeyr 414882 +85e2- = 272.5 8se2-(1.414)=272.5 8se2-=212.5=192.715 hence radius of anion will be 193pm 1.414