The number of lines exhibited by hydrogen(s) a is . The number of lines exhibite
ID: 963929 • Letter: T
Question
The number of lines exhibited by hydrogen(s) a is .
The number of lines exhibited by hydrogen(s) b is .
The number of lines exhibited by hydrogen(s) c is .
The number of lines exhibited by hydrogen(s) a is .
The number of lines exhibited by hydrogen(s) b is .
The number of lines exhibited by hydrogen(s) c is .
The number of lines exhibited by hydrogen(s) a is .
The number of lines exhibited by hydrogen(s) b is .
The number of lines exhibited by hydrogen(s) c is .
The number of lines exhibited by hydrogen(s) a is .
The number of lines exhibited by hydrogen(s) b is .
The number of lines exhibited by hydrogen(s) c is .
Explanation / Answer
Q.1: Ha is surrounded by 2 Hb atoms. Hence number of lines exhibited by Ha = 2 + 1 = 3 (triplet)
Hb is surrounded by 3 Ha atoms. Hence number of lines exhibited by Hb = 3 + 1 = 4 (quadriplet)
Hc is surrounded by 2 Hb atoms. Hence number of lines exhibited by Hc = 2 + 1 = 3 (triplet)
Q.2:
Ha is not surrounded by any other atom. Hence number of lines exhibited by Ha = 0 + 1 = 1 (singlet)
Hb is surrounded by 3+3 = 6 Hc atoms. Hence number of lines exhibited by Hb = 6 + 1 = 7 (heptet)
Hc is surrounded by 1 Hb atoms. Hence number of lines exhibited by Hc = 1 + 1 = 2 (doublet)
Q.3:
Ha is surrounded by 3 terminal methyl H atoms and 1 Hc atom. Hence number of lines exhibited by Ha = 3+1+1 = 5
Hb is surrounded by 1 Hc atom. Hence number of lines exhibited by Hb = 1 + 1 = 2 (douplet)
Hc is surrounded by 3+3 = 6 Hb atoms. Hence number of lines exhibited by Hc = 6 + 1 = 7 (heptet)
Q.4:
Ha is not surrounded by any other atom. Hence number of lines exhibited by Ha = 0 + 1 = 1 (singlet)
Hb is surrounded by 2 Hc atoms. Hence number of lines exhibited by Hb = 2 + 1 = 3 (triplet)
Hc is surrounded by 2 Hb atoms and 2 H atom of terminal CH2 - Cl group. Hence number of lines exhibited by Hc = 2 + 2 + 1 = 5 (pentet)