The ratio of Xe and F2 is 1:1
squre planer
Trigonal bipyramidal
3 dominant to 1 recessive
Asuming that the F1 generation is heterozygous for a single trait and that the F2 cross is of 2 F1 offspring. Ex. Aa X Aa the phenotypic ratio is 3:1 dominant to recessive. The genotypic ratio is 1:2:1 AA:Aa:aa.
The derivative of xe is e. The derivative of xe is exe-1.
No, because they have a full valence shell. Chemical reactions proceed because atoms wish to share or exchange electrons. If an atom's valence shell is full, they tend not to do this under normal circumstances.
The value of the bond angle in XeF2 is 180 degrees.
For monohybrid cross the genotype ratio in f2 generation would be 1:2:1 and phenotype ratio would be 3: 1
9:3:3:1 was the ratio of Mendel's f2 generation for the two factor cross.
it is ratio of mono hybrid cross found in f2 generation .this ratio is 3:1
F1: all tall F2: 1:3 short:tall
Xe belongs to the noble gas family so has 8 valence electrons...Xe => 5s25p6....... Two of these are bonded with fluorine. Thus it is left with 6 electronss i.e. 3 lone pairs.... So hybridization is sp3d ....the shape that should be =>Trigonal bipyramidal.... But it has 3 lone pairs on equatorial plane & 2 bond pairs on axial .....so final shape =>LINEAR...
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