For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
To know more about hybrisisation:
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First let us see what
kind of bonds are formed in the compound. By drawing the structure, we see that
the kind of bonds are:
N =- triple bond -= C –
O
<span>So there is only
single bond between C and O therefore the hybridization of C is sp.</span>
ANWERS ~
We know that :
1 cal (th) = 4.184 J
1 J = 0.2390057361 cal (th) , so :
•55.2 j to cal > 13.193116635 cal
•110 call > 460.24 joule
•65 kj > divide the energy value by 4.184
= 15.535 kilocalories calorie (IT)
——————
Converting form C to F > (F-32)*5/9Understand it better if we have Fahrenheit just add to the equation mentioned to find Celsius.
+to find F to C> (9/5*C)+32
•425 Fahrenheit = (425- 32) × 5/9 =218.33333333 Celsius
•1935 C = 3515 F
———————————-
Converting Celsius to kelvin,We know that :
K = C + 273.15
C = K - 273.15
And from F to K=9/5(F+459.67)
And K to F =(9/5 *k)-459.67
•39.4 Celsius = 312.55 kelvin
•337 Fahrenheit = (337+ 459.67) × 5/9 =442.594 kelvin
