In the electron configuration of tin (Sn), the 3d electrons are not present. Tin has a configuration of [Kr] 5s2 4d10 5p2, which means it has 2 electrons in the 5s orbital, 10 electrons in the 4d orbital, and 2 electrons in the 5p orbital.
There can only be 2 electrons in each single orbital, and they will be on opposite sides of the electron cloud (orbital).
Each planet in the solar system has a different orbital period, corresponding to the different sizes of their elliptical orbits.For the Earth, the present orbital period is 365.25636days. (rounded)
Sodium (Na) has 3 orbital boxes, corresponding to the s, p, and d orbitals. The s-orbital can hold up to 2 electrons, the p-orbital up to 6 electrons, and the d-orbital up to 10 electrons.
Four pairs of electrons, with one unpaired. This is the electronic configuration of fluorine
There are 3 nodes present in a 4f orbital: one radial node and two angular nodes. This means that there are regions in the orbital where the probability of finding an electron is zero.
There can be 10 electrons in the n=2 shell. Two can fit in the 1s orbital, two can fit in the 2s orbital, and six can fit in the 2p orbital.
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
An unoccupied orbital is represented by an empty box in orbital notation. It does not have any electrons present in it but is available to accept electrons if needed.
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
With greyhond
the value of the exponent n1
the value of the exponent n1
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2-1) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
In the electron configuration of tin (Sn), the 3d electrons are not present. Tin has a configuration of [Kr] 5s2 4d10 5p2, which means it has 2 electrons in the 5s orbital, 10 electrons in the 4d orbital, and 2 electrons in the 5p orbital.
five
void main() { int i; float n1,n2; abc: printf("Enter two nos "); scanf("%f%f",&n1,&n2); printf("\n %f + %f = %f " ,n1,n2,n1+n2); printf("\n %f - %f = %f " ,n1,n2,n1-n2); printf("\n %f x %f = %f " ,n1,n2,n1*n2); printf("\n %f / %f = %f " ,n1,n2,n1/n2); printf("\npress 5 to make another calculation"); scanf("%d",&i); if (i==5) goto abc; }