Arrange the following atoms in order of increasing effective nuclear charge experienced by the electrons in the \(n=2\) shell: Be, Br, Na, P, Se.

Find the atomic number of Be, Br, Na, P, and Se and arrange them in order. This can be done by referring to the periodic table. By referring to the periodic table, we have: Be (atomic number 4), Na (atomic number 11), P (atomic number 15), Se (atomic number 34), and Br (atomic number 35). Our initial sequence will be: Be - Na - P - Se - Br

Now, we will consider the electron shielding effect in the n=2 shell. In a multi-electron atom, the inner shell electrons tend to shield the outer shell electrons from the full influence of the nuclear charge. Since we are concerned with the n=2 shell, we only need to consider the shielding from the n=1 shell. For each element in our sequence, the n=1 shell is full, containing 2 electrons: Be (4: 2 electrons in n=1, 2 electrons in n=2) Na (11: 2 electrons in n=1, 8 electrons in n=2, 1 electron in n=3) P (15: 2 electrons in n=1, 8 electrons in n=2, 5 electrons in n=3) Se (34: 2 electrons in n=1, 8 electrons in n=2, 18 electrons in n=3, 6 electrons in n=4) Br (35: 2 electrons in n=1, 8 electrons in n=2, 18 electrons in n=3, 7 electrons in n=4) In all cases, the n=1 electrons provide shielding for the n=2 electrons.

The effective nuclear charge can be estimated using the formula: Z_eff = Z - S where Z is the atomic number and S is the shielding constant. For simplicity, we'll assume that each n=1 electron shields one unit of charge, so S = 2 for all elements. Be: Z_eff = 4 - 2 = 2 Na: Z_eff = 11 - 2 = 9 P: Z_eff = 15 - 2 = 13 Se: Z_eff = 34 - 2 = 32 Br: Z_eff = 35 - 2 = 33

Using the calculated effective nuclear charges, we can now arrange the elements in order of increasing Z_eff. Our final order is: Be (Z_eff = 2) - Na (Z_eff = 9) - P (Z_eff = 13) - Se (Z_eff = 32) - Br (Z_eff = 33) Thus, the atoms in order of increasing effective nuclear charge experienced by the electrons in the n=2 shell are Be, Na, P, Se, and Br.