Question

Atomic number of ag 47 in the same group that atomic numbers of elements placed above and below 18 long form of periodic table will be

Answer 1

Dear student

The atomic number of element above Ag = atomic number of Ag - 18 = 47 - 18 = 29

The atomic number of element below Ag = atomic number of Ag +32 = 47 +32 = 79

option (c) correct

from next tym wrote the question fully

Answer 2

Answer:

Ag with atomic number 47 is present in the d−block. Element above Ag is 47−18=29, element with atomic number 29,i.e.; Cu and element below Ag is 79 which is atomic number of Au

The element should have been 47+18=65 (Tb), but is present in lanthanides⟶ inner transition elements.