Question #c3e59

A. #"HOCH"_2"CH"_2"OH"#

This compound has 1 #"C-C"# bond, 2 #"C-O"# bonds, 4 #"C-H"# bond, and 2 #"O-H"# bonds, for 9 covalent bonds.

B. #"CH"_3"CH"_2"OH"#

Ethanol

This compound has 1 #"C-C"# bond, 1 #"C-O"# bond, 5 #"C-H"# bonds, and 1 #"O-H"# bond, for a total of eight covalent bonds.

C. #"HOCOCOOH"#

This compound contains 1 #"C-C"# bond, 6 #"C-O"# bonds (each #"C=O"# counts as two covalent bonds), and 2 #"O-H"# bonds, for 9 covalent bonds.

D. #"HOCH"_2"COOH"#

This compound contains 1 #"C-C"# bond, 4 #"C-O"# bonds, 2 #"C-H"# bonds, and 2 #"O-H"# bonds, for 9 covalent bonds.

We can use the following formula to determine the number of covalent bonds in a molecule.

#"Number of covalent bonds"=(x-y)/2#

where #x# is the total number of electrons required to fulfil the octet for atom other than hydrogen atom and doublet for hydrogen atom.

and #y# is the total number of valence electrons.

A) For compound #HOCH_2CH_2OH#

which consists of 2C ,2O and 6H ,we have

#x=2*8+2*8+6*2=44#

#y=2*4+2*6+6*1=26#

So #"Number of covalent bonds"=(x-y)/2=(44-26)/2=9#

B) For compound #CH_3CH_2OH#

which consists of 2C ,1O and 6H ,we have

#x=2*8+1*8+6*2=36#

#y=2*4+1*6+6*1=20#

So #"Number of covalent bonds"=(x-y)/2=(36-20)/2=8#

C) For compound #HOCOCOOH#

which consists of 2C ,4O and 2H ,we have

#x=2*8+4*8+2*2=52#

#y=2*4+4*6+2*1=34#

So #"Number of covalent bonds"=(x-y)/2=(52-34)/2=9#

D) For compound #HOCH_2COOH#

which consists of 2C ,3O and 4H ,we have

#x=2*8+3*8+4*2=48#

#y=2*4+3*6+4*1=30#

So #"Number of covalent bonds"=(x-y)/2=(48-30)/2=9#